A table that is partitioned by range is partitioned in such a way that each partition contains rows for which the partitioning expression value lies within a given range. Ranges should be contiguous but not overlapping, and are defined using the VALUES LESS THAN operator. For the next few examples, suppose that you are creating a table such as the following to hold personnel records for a chain of 20 video stores, numbered 1 through 20:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT NOT NULL,
    store_id INT NOT NULL
);

This table can be partitioned by range in a number of ways, depending on your needs. One way would be to use the store_id column. For instance, you might decide to partition the table 4 ways by adding a PARTITION BY RANGE clause as shown here:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT NOT NULL,
    store_id INT NOT NULL
)
PARTITION BY RANGE (store_id) (
    PARTITION p0 VALUES LESS THAN (6),
    PARTITION p1 VALUES LESS THAN (11),
    PARTITION p2 VALUES LESS THAN (16),
    PARTITION p3 VALUES LESS THAN (21)
);

In this partitioning scheme, all rows corresponding to employees working at stores 1 through 5 are stored in partition p0, to those employed at stores 6 through 10 are stored in partition p1, and so on. Note that each partition is defined in order, from lowest to highest. This is a requirement of the PARTITION BY RANGE syntax; you can think of it as being analogous to a switch ... case in C or Java in this regard.

It is easy to determine that a new row containing the data (72, 'Michael', 'Widenius', '1998-06-25', NULL, 13) is inserted into partition p2, but what happens when your chain adds a 21^st store? Under this scheme, there is no rule that covers a row whose store_id is greater than 20, so an error results because the server does not know where to place it. You can keep this from occurring by using a “catchall” VALUES LESS THAN clause in the CREATE TABLE statement that provides for all values greater than highest value explicitly named:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT NOT NULL,
    store_id INT NOT NULL
)
PARTITION BY RANGE (store_id) (
    PARTITION p0 VALUES LESS THAN (6),
    PARTITION p1 VALUES LESS THAN (11),
    PARTITION p2 VALUES LESS THAN (16),
    PARTITION p3 VALUES LESS THAN MAXVALUE
);

MAXVALUE represents the greatest possible integer value. Now, any rows whose store_id column value is greater than or equal to 16 (the highest value defined) are stored in partition p3. At some point in the future — when the number of stores has increased to 25, 30, or more — you can use an ALTER TABLE statement to add new partitions for stores 21-25, 26-30, and so on (see Section 16.3, “Partition Management”, for details of how to do this).

In much the same fashion, you could partition the table based on employee job codes — that is, based on ranges of job_code column values. For example — assuming that two-digit job codes are used for regular (in-store) workers, three-digit codes are used for office and support personnel, and four-digit codes are used for management positions — you could create the partitioned table using the following:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT NOT NULL,
    store_id INT NOT NULL
)
PARTITION BY RANGE (job_code) (
    PARTITION p0 VALUES LESS THAN (100),
    PARTITION p1 VALUES LESS THAN (1000),
    PARTITION p2 VALUES LESS THAN (10000)
);

In this instance, all rows relating to in-store workers would be stored in partition p0, those relating to office and support staff in p1, and those relating to managers in partition p2.

It is also possible to use an expression in VALUES LESS THAN clauses. However, MySQL must be able to evaluate the expression's return value as part of a LESS THAN (<) comparison.

Rather than splitting up the table data according to store number, you can use an expression based on one of the two DATE columns instead. For example, let us suppose that you wish to partition based on the year that each employee left the company; that is, the value of YEAR(separated). An example of a CREATE TABLE statement that implements such a partitioning scheme is shown here:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT,
    store_id INT
)
PARTITION BY RANGE ( YEAR(separated) ) (
    PARTITION p0 VALUES LESS THAN (1991),
    PARTITION p1 VALUES LESS THAN (1996),
    PARTITION p2 VALUES LESS THAN (2001),
    PARTITION p3 VALUES LESS THAN MAXVALUE
);

In this scheme, for all employees who left before 1991, the rows are stored in partition p0; for those who left in the years 1991 through 1995, in p1; for those who left in the years 1996 through 2000, in p2; and for any workers who left after the year 2000, in p3.

Range partitioning is particularly useful when:

List partitioning in MySQL is similar to range partitioning in many ways. As in partitioning by RANGE, each partition must be explicitly defined. The chief difference is that, in list partitioning, each partition is defined and selected based on the membership of a column value in one of a set of value lists, rather than in one of a set of contiguous ranges of values. This is done by using PARTITION BY LIST(expr) where expr is a column value or an expression based on a column value and returning an integer value, and then defining each partition by means of a VALUES IN (value_list), where value_list is a comma-separated list of integers.

Note: In MySQL 5.1, it is possible to match against only a list of integers (and possibly NULL — see Section 16.2.6, “How MySQL Partitioning Handles NULL Values”) when partitioning by LIST.

Unlike the case with partitions defined by range, list partitions do not need to be declared in any particular order. For more detailed syntactical information, see Section 13.1.10, “CREATE TABLE Syntax”.

For the examples that follow, we assume that the basic definition of the table to be partitioned is provided by the CREATE TABLE statement shown here:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT,
    store_id INT
);

(This is the same table used as a basis for the examples in Section 16.2.1, “RANGE Partitioning”.)

Suppose that there are 20 video stores distributed among 4 franchises as shown in the following table:

To partition this table in such a way that rows for stores belonging to the same region are stored in the same partition, you could use the CREATE TABLE statement shown here:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT,
    store_id INT
)
PARTITION BY LIST(store_id) (
    PARTITION pNorth VALUES IN (3,5,6,9,17),
    PARTITION pEast VALUES IN (1,2,10,11,19,20),
    PARTITION pWest VALUES IN (4,12,13,14,18),
    PARTITION pCentral VALUES IN (7,8,15,16)
);

This makes it easy to add or drop employee records relating to specific regions to or from the table. For instance, suppose that all stores in the West region are sold to another company. All rows relating to employees working at stores in that region can be deleted with the query ALTER TABLE employees DROP PARTITION pWest;, which can be executed much more efficiently than the equivalent DELETE statement DELETE FROM employees WHERE store_id IN (4,12,13,14,18);.

As with RANGE and HASH partitioning, if you wish to partition a table by a column whose value is not an integer or NULL, you must employ a partitioning expression based on that column which returns such a value. For example, suppose that the table containing employee data is defined as shown here:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code CHAR(1),
    store_id INT
);

In this version of the employees table, the job code is a letter rather than a number. Each letter corresponds to a specific job, and we wish to partition the table in such a way that records for employees having similar jobs or working in the same department are grouped into the same partition, according to the following scheme:

Since we cannot use character values in value-lists, we need to convert these into integers or NULLs. For this purpose, we can use the ASCII() function on the column value. A partitioned table that implements this scheme is shown here:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code CHAR(1),
    store_id INT
) 
PARTITION BY LIST( ASCII(job_code) ) (
    PARTITION management VALUES IN(68, 77, 79, 80),
    PARTITION sales VALUES IN(66, 76, 83),
    PARTITION technical VALUES IN(65, 69, 71, 73, 84),
    PARTITION clerical VALUES IN(75, 78, 89),
    PARTITION support VALUES IN(67, 70, 74, 82, 86),
    PARTITION unassigned VALUES IN(NULL, 0, 32)
);

Since expressions are not permitted in partition value lists, you must list the ASCII codes for the letters that are to be matched. Note that ASCII(NULL) returns NULL.

Important: If you try to insert a row such that the column value (or the partitioning expression's return value) is not found in any of the partitioning value lists, the INSERT query will fail with an error. For example, given the LIST partitioning scheme just outlined, this query will fail:

INSERT INTO employees VALUES 
    (224, 'Linus', 'Torvalds', '2002-05-01', '2004-10-12', 'Q', 21);

Failure occurs because 81 (the ASCII code for the uppercase letter 'Q' is not found in any of the value lists used to define any of the partitions. There is no “catch-all” definition for list partitions analogous to VALUES LESS THAN(MAXVALUE) which accommodates values not found in any of the value lists. In other words, any value which is to be matched must be found in one of the value lists.

As with RANGE partitioning, it is possible to combine LIST partitioning with partitioning by hash or key to produce a composite partitioning (subpartitioning). See Section 16.2.5, “Subpartitioning”.

Partitioning by HASH is used primarily to ensure an even distribution of data among a predetermined number of partitions. With range or list partitioning, you must specify explicitly into which partition a given column value or set of column values is to be stored; with hash partitioning, MySQL takes care of this for you, and you need only specify a column value or expression based on a column value to be hashed and the number of partitions into which the partitioned table is to be divided.

To partition a table using HASH partitioning, it is necessary to append to the CREATE TABLE statement a PARTITION BY HASH (expr) clause, where expr is an expression that returns an integer. This can simply be the name of a column whose type is one of MySQL's integer types. In addition, you will most likely want to follow this with a PARTITIONS num clause, where num is a non-negative integer representing the number of partitions into which the table is to be divided.

For example, the following statement creates a table that uses hashing on the store_id column and is divided into 4 partitions:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT,
    store_id INT
)
PARTITION BY HASH(store_id)
PARTITIONS 4;

If you do not include a PARTITIONS clause, the number of partitions defaults to 1.

Using the PARTITIONS keyword without a number following it results in a syntax error.

You can also use an SQL expression that returns an integer for expr. For instance, you might want to partition based on the year in which an employee was hired. This can be done as shown here:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT,
    store_id INT
)
PARTITION BY HASH( YEAR(hired) )
PARTITIONS 4;

You may use any function or other expression for expr that is valid in MySQL, so long as it returns a non-constant, non-random integer value. (In other words, it should be varying but deterministic.) However, you should keep in mind that this expression is evaluated each time a row is inserted or updated (or possibly deleted); this means that very complex expressions may give rise to performance issues, particularly when performing operations (such as batch inserts) that affect a great many rows at one time.

The most efficient hashing function is one which operates upon a single table column and whose value increases or decreases consistently with the column value, as this allows for “pruning” on ranges of partitions. That is, the more closely that the expression varies with the value of the column on which it is based, the more efficiently MySQL can use the expression for hash partitioning.

For example, where date_col is a column of type DATE, then the expression TO_DAYS(date_col) is said to vary directly with the value of date_col, because for every change in the value of date_col, the value of the expression changes in a consistent manner. The variance of the expression YEAR(date_col) with respect to date_col is not quite as direct as that of TO_DAYS(date_col), because not every possible change in date_col produces an equivalent change in YEAR(date_col). Even so, YEAR(date_col) is a good candidate for a hashing function, because it varies directly with a portion of date_col and there is no possible change in date_col that produces a disproportionate change in YEAR(date_col).

By way of contrast, suppose that you have a column named int_col whose type is INT. Now consider the expression POW(5-int_col,3) + 6. This would be a poor choice for a hashing function because a change in the value of int_col is not guaranteed to produce a proportional change in the value of the expression. Changing the value of int_col by a given amount can produce by widely different changes in the value of the expression. For example, changing int_col from 5 to 6 produces a change of -1 in the value of the expression, but changing the value of int_col from 6 to 7 produces a change of -7 in the expression value.

In other words, the more closely the graph of the column value versus the value of the expression follows a straight line as traced by the equation y=nx where n is some nonzero constant, the better the expression is suited to hashing. This has to do with the fact that the more nonlinear an expression is, the more uneven the distribution of data among the partitions it tends to produce.

In theory, pruning is also possible for expressions involving more than one column value, but determining which of such expressions are suitable can be quite difficult and time-consuming. For this reason, the use of hashing expressions involving multiple columns is not particularly recommended.

When PARTITION BY HASH is used, MySQL determines which partition of num partitions to use based on the modulus of the result of the user function. In other words, for an expression expr, the partition in which the record is stored is partition number N, where N = MOD(expr, num). For example, suppose table t1 is defined as follows, so that it has 4 partitions:

CREATE TABLE t1 (col1 INT, col2 CHAR(5), col3 DATE)
    PARTITION BY HASH( YEAR(col3) )
    PARTITIONS 4;

If you insert a record into t1 whose col3 value is '2005-09-15', then the partition in which it is stored is determined as follows:

MOD(YEAR('2005-09-01'),4)
=  MOD(2005,4)
=  1

MySQL 5.1 also supports a variant of HASH partitioning known as linear hashing which employs a more complex algorithm for determining the placement of new rows inserted into the partitioned table. See Section 16.2.3.1, “LINEAR HASH Partitioning”, for a description of this algorithm.

The user function is evaluated each time a record is inserted or updated. It may also — depending on the circumstances — be evaluated when records are deleted.

Note: If the table to be partitioned has a UNIQUE key, then any columns supplied as arguments to the HASH user function or to the KEY's column_list must be part of that key.

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT,
    store_id INT
)
PARTITION BY LINEAR HASH( YEAR(hired) )
PARTITIONS 4;

CREATE TABLE t1 (col1 INT, col2 CHAR(5), col3 DATE)
    PARTITION BY LINEAR HASH( YEAR(col3) )
    PARTITIONS 6;

V = POWER(2, CEILING( LOG(2,7) )) = 8
N = YEAR('2003-04-14') & (8 - 1)
   = 2003 & 7
   = 3

(3 >= 6 is FALSE: record stored in partition #3)

V = 8
N = YEAR('1998-10-19') & (8-1)
  = 1998 & 7
  = 6

(6 >= 6 is TRUE: additional step required)

N = 6 & CEILING(5 / 2)
  = 6 & 3
  = 2

(2 >= 6 is FALSE: record stored in partition #2)

Partitioning by key is similar to partitioning by hash, except that where hash partitioning employs a user-defined expression, the hashing function for key partitioning is supplied by the MySQL server. MySQL Cluster uses MD5() for this purpose; for tables using other storage engines, the server employs its own internal hashing function which is based on the same algorithm as PASSWORD().

The syntax rules for CREATE TABLE ... PARTITION BY KEY are similar to those for creating a table that is partitioned by hash. The major differences are that:

It is also possible to partition a table by linear key. Here is a simple example:

CREATE TABLE tk (
    col1 INT NOT NULL,
    col2 CHAR(5),
    col3 DATE
) 
PARTITION BY LINEAR KEY (col1)
PARTITIONS 3;

Using LINEAR has the same effect on KEY partitioning as it does on HASH partitioning, with the partition number being derived using a powers-of-two algorithm rather than modulo arithmetic. See Section 16.2.3.1, “LINEAR HASH Partitioning”, for a description of this algorithm and its implications.

Subpartitioning — also known as composite partitioning — is the further division of each partition in a partitioned table. For example, consider the following CREATE TABLE statement:

CREATE TABLE ts (id INT, purchased DATE)
    PARTITION BY RANGE( YEAR(purchased) )
    SUBPARTITION BY HASH( TO_DAYS(purchased) )
    SUBPARTITIONS 2 (
        PARTITION p0 VALUES LESS THAN (1990),
        PARTITION p1 VALUES LESS THAN (2000),
        PARTITION p2 VALUES LESS THAN MAXVALUE
    );

Table ts has 3 RANGE partitions. Each of these partitions — p0, p1, and p2 — is further divided into 2 subpartitions. In effect, the entire table is divided into 3 * 2 = 6 partitions. However, due to the action of the PARTITION BY RANGE clause, the first 2 of these store only those records with a value less than 1990 in the purchased column.

In MySQL 5.1, it is possible to subpartition tables that are partitioned by RANGE or LIST. Subpartitions may use either HASH or KEY partitioning. This is also known as composite partitioning.

It is also possible to define subpartitions explicitly using SUBPARTITION clauses to specify options for individual subpartitions. For example, a more verbose fashion of creating the same table ts as shown in the previous example would be:

CREATE TABLE ts (id INT, purchased DATE)
    PARTITION BY RANGE( YEAR(purchased) )
    SUBPARTITION BY HASH( TO_DAYS(purchased) ) (
        PARTITION p0 VALUES LESS THAN (1990) (
            SUBPARTITION s0,
            SUBPARTITION s1
        ),
        PARTITION p1 VALUES LESS THAN (2000) (
            SUBPARTITION s2,
            SUBPARTITION s3
        ),
        PARTITION p2 VALUES LESS THAN MAXVALUE (
            SUBPARTITION s4,
            SUBPARTITION s5
        )
    );

Some syntactical items of note:

Subpartitions can be used with especially large tables to distribute data and indexes across many disks. Suppose that you have 6 disks mounted as /disk0, /disk1, /disk2, and so on. Now consider the following example:

CREATE TABLE ts (id INT, purchased DATE)
    PARTITION BY RANGE( YEAR(purchased) )
    SUBPARTITION BY HASH( TO_DAYS(purchased) ) (
        PARTITION p0 VALUES LESS THAN (1990) (
            SUBPARTITION s0 
                DATA DIRECTORY = '/disk0/data' 
                INDEX DIRECTORY = '/disk0/idx',
            SUBPARTITION s1 
                DATA DIRECTORY = '/disk1/data' 
                INDEX DIRECTORY = '/disk1/idx'
        ),
        PARTITION p1 VALUES LESS THAN (2000) (
            SUBPARTITION s2 
                DATA DIRECTORY = '/disk2/data' 
                INDEX DIRECTORY = '/disk2/idx',
            SUBPARTITION s3 
                DATA DIRECTORY = '/disk3/data' 
                INDEX DIRECTORY = '/disk3/idx'
        ),
        PARTITION p2 VALUES LESS THAN MAXVALUE (
            SUBPARTITION s4 
                DATA DIRECTORY = '/disk4/data' 
                INDEX DIRECTORY = '/disk4/idx',
            SUBPARTITION s5 
                DATA DIRECTORY = '/disk5/data' 
                INDEX DIRECTORY = '/disk5/idx'
        )
    );

In this case, a separate disk is used for the data and for the indexes of each RANGE. Many other variations are possible; another example might be:

CREATE TABLE ts (id INT, purchased DATE)
    PARTITION BY RANGE(YEAR(purchased))
    SUBPARTITION BY HASH( TO_DAYS(purchased) ) (
        PARTITION p0 VALUES LESS THAN (1990) (
            SUBPARTITION s0a 
                DATA DIRECTORY = '/disk0' 
                INDEX DIRECTORY = '/disk1',
            SUBPARTITION s0b 
                DATA DIRECTORY = '/disk2' 
                INDEX DIRECTORY = '/disk3'
        ),
        PARTITION p1 VALUES LESS THAN (2000) (
            SUBPARTITION s1a 
                DATA DIRECTORY = '/disk4/data' 
                INDEX DIRECTORY = '/disk4/idx',
            SUBPARTITION s1b 
                DATA DIRECTORY = '/disk5/data' 
                INDEX DIRECTORY = '/disk5/idx'
        ),
        PARTITION p2 VALUES LESS THAN MAXVALUE (
            SUBPARTITION s2a,
            SUBPARTITION s2b
        )
    );

Here, the storage is as follows:

Beginning with MySQL 5.1.18, the DATA DIRECTORY and INDEX DIRECTORY options are disallowed when the NO_DIR_IN_CREATE server SQL mode is in effect. This is true for partitions and subpartitions.

Partitioning in MySQL does nothing to disallow NULL as the value of a partitioning expression, whether it is a column value or the value of a user-supplied expression. Even though it is permitted to use NULL as the value of an expression that must otherwise yield an integer, it is important to keep in mind that NULL is not a number. Beginning version 5.1.8, MySQL Partitioning treats NULL as being less than any non-NULL value, just as ORDER BY does.

Because of this, this treatment of NULL varies between partitioning of different types, and may produce behavior which you do not expect if you are not prepared for it. This being the case, we discuss in this section how each MySQL partitioning types handles NULL values when determining the partition in which a row should be stored, and provide examples for each.

If you insert a row into a table partitioned by RANGE such that the column value used to determine the partition is NULL, the row is inserted into the lowest partition. For example, consider these two tables, created and populated as follows:

mysql> CREATE TABLE t1 (
    ->     c1 INT, 
    ->     c2 VARCHAR(20)
    -> ) 
    -> PARTITION BY RANGE(c1) ( 
    ->     PARTITION p0 VALUES LESS THAN (0), 
    ->     PARTITION p1 VALUES LESS THAN (10), 
    ->     PARTITION p2 VALUES LESS THAN MAXVALUE 
    -> );
Query OK, 0 rows affected (0.09 sec)
 
mysql> CREATE TABLE t1 (
    ->     c1 INT, 
    ->     c2 VARCHAR(20)
    -> ) 
    -> PARTITION BY RANGE(c1) ( 
    ->     PARTITION p0 VALUES LESS THAN (-5), 
    ->     PARTITION p1 VALUES LESS THAN (0),  
    ->     PARTITION p1 VALUES LESS THAN (10), 
    ->     PARTITION p2 VALUES LESS THAN MAXVALUE 
    -> );
Query OK, 0 rows affected (0.09 sec)

mysql> INSERT INTO t1 VALUES (NULL, 'mothra');
Query OK, 1 row affected (0.00 sec)

mysql> INSERT INTO t2 VALUES (NULL, 'mothra');
Query OK, 1 row affected (0.00 sec)

mysql> SELECT * FROM t1;
+------+--------+
| id   | name   |
+------+--------+
| NULL | mothra |
+------+--------+
1 row in set (0.00 sec)

mysql> SELECT * FROM t2;
+------+--------+
| id   | name   |
+------+--------+
| NULL | mothra |
+------+--------+
1 row in set (0.00 sec)

You can see which partitions the rows are stored in by inspecting the filesystem and comparing the sizes of the .MYD files correpsonding to the partitions:

/var/lib/mysql/test> ls -l *.MYD
-rw-rw----  1 mysql mysql 20 2006-03-10 03:27 t1#P#p0.MYD
-rw-rw----  1 mysql mysql  0 2006-03-10 03:17 t1#P#p1.MYD
-rw-rw----  1 mysql mysql  0 2006-03-10 03:17 t1#P#p2.MYD
-rw-rw----  1 mysql mysql 20 2006-03-10 03:27 t2#P#p0.MYD
-rw-rw----  1 mysql mysql  0 2006-03-10 03:17 t2#P#p1.MYD
-rw-rw----  1 mysql mysql  0 2006-03-10 03:17 t2#P#p2.MYD
-rw-rw----  1 mysql mysql  0 2006-03-10 03:17 t2#P#p3.MYD

(Partition files are named according to the format table_name#P#partition_name.extension, so that t1#P#p0.MYD is the file in which data belonging to partition p0 of table t1 is stored. Note: Prior to MySQL 5.1.5, these files would have been named t1_p0.MYD and t2_p0.MYD, respectively. See Section C.1.21, “Changes in release 5.1.6 (01 February 2006)” and Bug#13437 for information regarding how this change impacts upgrades.)

You can also demonstrate that these rows were stored in the lowest partition of the each table by dropping these partitions, and then re-running the SELECT statements:

mysql> ALTER TABLE t1 DROP PARTITION p0;
Query OK, 0 rows affected (0.16 sec)

mysql> ALTER TABLE t2 DROP PARTITION p0;
Query OK, 0 rows affected (0.16 sec)

mysql> SELECT * FROM t1;
Empty set (0.00 sec)

mysql> SELECT * FROM t2;
Empty set (0.00 sec)

(For more information on ALTER TABLE ... DROP PARTITION, see Section 13.1.4, “ALTER TABLE Syntax”.)

Such treatment also holds true for partitioning expressions that use SQL functions. Suppose that we have a table such as this one:

CREATE TABLE tndate (
    id INT,
    dt DATE
)
PARTITION BY RANGE( YEAR(dt) ) (
    PARTITION p0 VALUES LESS THAN (1990),
    PARTITION p1 VALUES LESS THAN (2000),
    PARTITION p2 VALUES LESS THAN MAXVALUE
);

As with other MySQL functions, YEAR(NULL) returns NULL. A row with a dt column value of NULL is treated as though the partitioning expression evaluated to a value less than any other value, and so is inserted into partition p0.

A table that is partitioned by LIST admits NULL values if and only if one of its partitions is defined using that value-list that contains NULL. The converse of this is that a table partitioned by LIST which does not explicitly use NULL in a value list rejects rows resulting in a NULL value for the partitioning expression, as shown in this example:

mysql> CREATE TABLE ts1 (
    ->     c1 INT,
    ->     c2 VARCHAR(20)
    -> )
    -> PARTITION BY LIST(c1) (
    ->     PARTITION p0 VALUES IN (0, 3, 6),
    ->     PARTITION p1 VALUES IN (1, 4, 7),
    ->     PARTITION p2 VALUES IN (2, 5, 8)
    -> );
Query OK, 0 rows affected (0.01 sec)

mysql> INSERT INTO ts1 VALUES (9, 'mothra');
ERROR 1504 (HY000): Table has no partition for value 9

mysql> INSERT INTO ts1 VALUES (NULL, 'mothra');
ERROR 1504 (HY000): Table has no partition for value NULL

Only rows having a c1 value between 0 and 8 inclusive can be inserted into ts1. NULL falls outside this range, just like the number 9. We can create tables ts2 and ts3 having value lists containing NULL, as shown here:

mysql> CREATE TABLE ts2 (
    ->     c1 INT,
    ->     c2 VARCHAR(20)
    -> )
    -> PARTITION BY LIST(c1) (
    ->     PARTITION p0 VALUES IN (0, 3, 6),
    ->     PARTITION p1 VALUES IN (1, 4, 7),
    ->     PARTITION p2 VALUES IN (2, 5, 8),
    ->     PARTITION p3 VALUES IN (NULL)
    -> );
Query OK, 0 rows affected (0.01 sec)

mysql> CREATE TABLE ts3 (
    ->     c1 INT,
    ->     c2 VARCHAR(20)
    -> )
    -> PARTITION BY LIST(c1) (
    ->     PARTITION p0 VALUES IN (0, 3, 6),
    ->     PARTITION p1 VALUES IN (1, 4, 7, NULL),
    ->     PARTITION p2 VALUES IN (2, 5, 8)
    -> );
Query OK, 0 rows affected (0.01 sec)

When defining value lists for partitioning, you can treat NULL just as you would any other value, and so VALUES IN (NULL) and VALUES IN (1, 4, 7, NULL) are both valid (as are VALUES IN (1, NULL, 4, 7), VALUES IN (NULL, 1, 4, 7), and so on). You can insert a row having NULL for column c1 into each of the tables ts2 and ts3:

mysql> INSERT INTO ts2 VALUES (NULL, 'mothra');
Query OK, 1 row affected (0.00 sec)

mysql> INSERT INTO ts3 VALUES (NULL, 'mothra');
Query OK, 1 row affected (0.00 sec)

By inspecting the filesystem, you can verify that the first of these statements inserted a new row into partition p3 of table ts2, and that the second statement inserted a new row into partition p1 of table ts3:

/var/lib/mysql/test> ls -l ts2*.MYD
-rw-rw----  1 mysql mysql  0 2006-03-10 10:35 ts2#P#p0.MYD
-rw-rw----  1 mysql mysql  0 2006-03-10 10:35 ts2#P#p1.MYD
-rw-rw----  1 mysql mysql  0 2006-03-10 10:35 ts2#P#p2.MYD
-rw-rw----  1 mysql mysql 20 2006-03-10 10:35 ts2#P#p3.MYD

/var/lib/mysql/test> ls -l ts3*.MYD
-rw-rw----  1 mysql mysql  0 2006-03-10 10:36 ts3#P#p0.MYD
-rw-rw----  1 mysql mysql 20 2006-03-10 10:36 ts3#P#p1.MYD
-rw-rw----  1 mysql mysql  0 2006-03-10 10:36 ts3#P#p2.MYD

As in earlier examples, we assume the use of the bash shell on a Unix operating system for listing files; use whatever your platform provides in this regard. For example, if you are using a DOS shell on a Windows operating system, the equivalent for the last listing might be obtained by running the command dir ts3*.MYD in the directory C:\Program Files\MySQL\MySQL Server 5.1\data\test.

As shown earlier in this section, you can also verify which partitions were used for storing the values by deleting them and then performing a SELECT.

NULL is handled somewhat differently for tables partitioned by HASH or KEY. In these cases, any partition expression that yields a NULL value is treated as though its return value were zero. We can verify this behavior by examining the effects on the filesystem of creating a table partitioned by HASH and populating it with a record containing appropriate values. Suppose that you have a table th, created in the test database, using this statement:

mysql> CREATE TABLE th (
    ->     c1 INT,
    ->     c2 VARCHAR(20)
    -> )
    -> PARTITION BY HASH(c1)
    -> PARTITIONS 2;
Query OK, 0 rows affected (0.00 sec)

Assuming an RPM installation of MySQL on Linux, this statement creates two .MYD files in /var/lib/mysql/test, which can be viewed in the bash shell as follows:

/var/lib/mysql/test> ls th*.MYD -l
-rw-rw----  1 mysql mysql 0 2005-11-04 18:41 th#P#p0.MYD
-rw-rw----  1 mysql mysql 0 2005-11-04 18:41 th#P#p1.MYD

Note that the size of each file is 0 bytes. Now insert a row into th whose c1 column value is NULL, and verify that this row was inserted:

mysql> INSERT INTO th VALUES (NULL, 'mothra');
Query OK, 1 row affected (0.00 sec)

mysql> SELECT * FROM th;
+------+---------+
| c1   | c2      |
+------+---------+
| NULL | mothra  |
+------+---------+
1 row in set (0.01 sec)

Recall that for any integer N, the value of NULL MOD N is always NULL. For tables that are partitioned by HASH or KEY, this result is treated for determining the correct partition as 0. Returning to the system shell (still assuming bash for this purpose), we can see that the value was inserted into the first partition (named p0 by default) by listing the data files once again:

var/lib/mysql/test> ls *.MYD -l
-rw-rw----  1 mysql mysql 20 2005-11-04 18:44 th#P#p0.MYD
-rw-rw----  1 mysql mysql  0 2005-11-04 18:41 th#P#p1.MYD

You can see that the INSERT statement modified only the file th#P#p0.MYD (increasing its size on disk), without affecting the other data file.

Important: Prior to MySQL 5.1.8, RANGE partitioning treated a partitioning expression value of NULL as a zero with respect to determining placement (the only way to circumvent this was to design tables so as not to allow nulls, usually by declaring columns NOT NULL). If you have a RANGE partitioning scheme that depends on this earlier behavior, you will need to re-implement it when upgrading to MySQL 5.1.8 or later.

Range and list partitions are very similar with regard to how the adding and dropping of partitions are handled. For this reason we discuss the management of both sorts of partitioning in this section. For information about working with tables that are partitioned by hash or key, see Section 16.3.2, “Management of HASH and KEY Partitions”. Dropping a RANGE or LIST partition is more straightforward than adding one, so we discuss this first.

Dropping a partition from a table that is partitioned by either RANGE or by LIST can be accomplished using the ALTER TABLE statement with a DROP PARTITION clause. Here is a very basic example, which supposes that you have already created a table which is partitioned by range and then populated with 10 records using the following CREATE TABLE and INSERT statements:

mysql> CREATE TABLE tr (id INT, name VARCHAR(50), purchased DATE)
    ->     PARTITION BY RANGE( YEAR(purchased) ) (
    ->         PARTITION p0 VALUES LESS THAN (1990),
    ->         PARTITION p1 VALUES LESS THAN (1995),
    ->         PARTITION p2 VALUES LESS THAN (2000),
    ->         PARTITION p3 VALUES LESS THAN (2005)
    ->     );
Query OK, 0 rows affected (0.01 sec)

mysql> INSERT INTO tr VALUES
    ->     (1, 'desk organiser', '2003-10-15'),
    ->     (2, 'CD player', '1993-11-05'),
    ->     (3, 'TV set', '1996-03-10'),
    ->     (4, 'bookcase', '1982-01-10'),
    ->     (5, 'exercise bike', '2004-05-09'),
    ->     (6, 'sofa', '1987-06-05'),
    ->     (7, 'popcorn maker', '2001-11-22'),
    ->     (8, 'aquarium', '1992-08-04'),
    ->     (9, 'study desk', '1984-09-16'),
    ->     (10, 'lava lamp', '1998-12-25');
Query OK, 10 rows affected (0.01 sec)                  

You can see which items should have been inserted into partition p2 as shown here:

mysql> SELECT * FROM tr
    -> WHERE purchased BETWEEN '1995-01-01' AND '1999-12-31';
+------+-----------+------------+
| id   | name      | purchased  |
+------+-----------+------------+
|    3 | TV set    | 1996-03-10 |
|   10 | lava lamp | 1998-12-25 |
+------+-----------+------------+
2 rows in set (0.00 sec)

To drop the partition named p2, execute the following command:

mysql> ALTER TABLE tr DROP PARTITION p2;
Query OK, 0 rows affected (0.03 sec)

Note: In MySQL 5.1, the NDBCLUSTER storage engine does not support ALTER TABLE ... DROP PARTITION. It does, however, support the other partitioning-related extensions to ALTER TABLE that are described in this chapter.

It is very important to remember that, when you drop a partition, you also delete all the data that was stored in that partition. You can see that this is the case by re-running the previous SELECT query:

mysql> SELECT * FROM tr WHERE purchased 
    -> BETWEEN '1995-01-01' AND '1999-12-31';
Empty set (0.00 sec)

Because of this, the requirement was added in MySQL 5.1.10 that you have the DROP privilege for a table before you can execute ALTER TABLE ... DROP PARTITION on that table.

If you wish to drop all data from all partitions while preserving the table definition and its partitioning scheme, use the TRUNCATE TABLE command. (See Section 13.2.9, “TRUNCATE Syntax”.)

If you intend to change the partitioning of a table without losing data, use ALTER TABLE ... REORGANIZE PARTITION instead. See below or in Section 13.1.4, “ALTER TABLE Syntax”, for information about REORGANIZE PARTITION.

If you now execute a SHOW CREATE TABLE command, you can see how the partitioning makeup of the table has been changed:

mysql> SHOW CREATE TABLE tr\G
*************************** 1. row ***************************
       Table: tr
Create Table: CREATE TABLE `tr` (
  `id` int(11) default NULL,
  `name` varchar(50) default NULL,
  `purchased` date default NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1 
PARTITION BY RANGE ( YEAR(purchased) ) (
  PARTITION p0 VALUES LESS THAN (1990) ENGINE = MyISAM, 
  PARTITION p1 VALUES LESS THAN (1995) ENGINE = MyISAM, 
  PARTITION p3 VALUES LESS THAN (2005) ENGINE = MyISAM
)
1 row in set (0.01 sec)

When you insert new rows into the changed table with purchased column values between '1995-01-01' and '2004-12-31' inclusive, those rows will be stored in partition p3. You can verify this as follows:

mysql> INSERT INTO tr VALUES (11, 'pencil holder', '1995-07-12');
Query OK, 1 row affected (0.00 sec)

mysql> SELECT * FROM tr WHERE purchased 
    -> BETWEEN '1995-01-01' AND '2004-12-31';
+------+----------------+------------+
| id   | name           | purchased  |
+------+----------------+------------+
|   11 | pencil holder  | 1995-07-12 |
|    1 | desk organiser | 2003-10-15 |
|    5 | exercise bike  | 2004-05-09 |
|    7 | popcorn maker  | 2001-11-22 |
+------+----------------+------------+
4 rows in set (0.00 sec)

mysql> ALTER TABLE tr DROP PARTITION p3;
Query OK, 0 rows affected (0.03 sec)

mysql> SELECT * FROM tr WHERE purchased 
    -> BETWEEN '1995-01-01' AND '2004-12-31';
Empty set (0.00 sec)

Note that the number of rows dropped from the table as a result of ALTER TABLE ... DROP PARTITION is not reported by the server as it would be by the equivalent DELETE query.

Dropping LIST partitions uses exactly the same ALTER TABLE ... DROP PARTITION syntax as used for dropping RANGE partitions. However, there is one important difference in the effect this has on your use of the table afterward: You can no longer insert into the table any rows having any of the values that were included in the value list defining the deleted partition. (See Section 16.2.2, “LIST Partitioning”, for an example.)

To add a new range or list partition to a previously partitioned table, use the ALTER TABLE ... ADD PARTITION statement. For tables which are partitioned by RANGE, this can be used to add a new range to the end of the list of existing partitions. For example, suppose that you have a partitioned table containing membership data for your organisation, which is defined as follows:

CREATE TABLE members (
    id INT, 
    fname VARCHAR(25),
    lname VARCHAR(25), 
    dob DATE
)
PARTITION BY RANGE( YEAR(dob) ) (
    PARTITION p0 VALUES LESS THAN (1970),
    PARTITION p1 VALUES LESS THAN (1980),
    PARTITION p2 VALUES LESS THAN (1990)
);

Suppose further that the minimum age for members is 16. As the calendar approaches the end of 2005, you realize that you will soon be admitting members who were born in 1990 (and later in years to come). You can modify the members table to accommodate new members born in the years 1990-1999 as shown here:

ALTER TABLE ADD PARTITION (PARTITION p3 VALUES LESS THAN (2000));

Important: With tables that are partitioned by range, you can use ADD PARTITION to add new partitions to the high end of the partitions list only. Trying to add a new partition in this manner between or before existing partitions will result in an error as shown here:

mysql> ALTER TABLE members
     >     ADD PARTITION (
     >     PARTITION p3 VALUES LESS THAN (1960));
ERROR 1463 (HY000): VALUES LESS THAN value must be strictly »
   increasing for each partition

In a similar fashion, you can add new partitions to a table that is partitioned by LIST. For example, given a table defined like so:

CREATE TABLE tt (
    id INT, 
    data INT
)
PARTITION BY LIST(data) (
    PARTITION p0 VALUES IN (5, 10, 15),
    PARTITION p1 VALUES IN (6, 12, 18)
);

You can add a new partition in which to store rows having the data column values 7, 14, and 21 as shown:

ALTER TABLE tt ADD PARTITION (PARTITION p2 VALUES IN (7, 14, 21));

Note that you cannot add a new LIST partition encompassing any values that are already included in the value list of an existing partition. If you attempt to do so, an error will result:

mysql> ALTER TABLE tt ADD PARTITION 
     >     (PARTITION np VALUES IN (4, 8, 12));
ERROR 1465 (HY000): Multiple definition of same constant »
                    in list partitioning

Because any rows with the data column value 12 have already been assigned to partition p1, you cannot create a new partition on table tt that includes 12 in its value list. To accomplish this, you could drop p1, and add np and then a new p1 with a modified definition. However, as discussed earlier, this would result in the loss of all data stored in p1 — and it is often the case that this is not what you really want to do. Another solution might appear to be to make a copy of the table with the new partitioning and to copy the data into it using CREATE TABLE ... SELECT ..., then drop the old table and rename the new one, but this could be very time-consuming when dealing with a large amounts of data. This also might not be feasible in situations where high availability is a requirement.

Beginning with MySQL 5.1.6, you can add multiple partitions in a single ALTER TABLE ... ADD PARTITION statement as shown here:

CREATE TABLE employees (
  id INT NOT NULL,
  fname VARCHAR(50) NOT NULL,
  lname VARCHAR(50) NOT NULL,
  hired DATE NOT NULL
)
PARTITION BY RANGE( YEAR(hired) ) (
  PARTITION p1 VALUES LESS THAN (1991),
  PARTITION p2 VALUES LESS THAN (1996),
  PARTITION p3 VALUES LESS THAN (2001),
  PARTITION p4 VALUES LESS THAN (2005)
);

ALTER TABLE employees ADD PARTITION (
    PARTITION p5 VALUES LESS THAN (2010), 
    PARTITION p6 VALUES LESS THAN MAXVALUE
);

Fortunately, MySQL's partitioning implementation provides ways to redefine partitions without losing data. Let us look first at a couple of simple examples involving RANGE partitioning. Recall the members table which is now defined as shown here:

mysql> SHOW CREATE TABLE members\G
*************************** 1. row ***************************
       Table: members
Create Table: CREATE TABLE `members` (
  `id` int(11) default NULL,
  `fname` varchar(25) default NULL,
  `lname` varchar(25) default NULL,
  `dob` date default NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1 
PARTITION BY RANGE ( YEAR(dob) ) (
  PARTITION p0 VALUES LESS THAN (1970) ENGINE = MyISAM, 
  PARTITION p1 VALUES LESS THAN (1980) ENGINE = MyISAM, 
  PARTITION p2 VALUES LESS THAN (1990) ENGINE = MyISAM.
  PARTITION p3 VALUES LESS THAN (2000) ENGINE = MyISAM
)

Suppose that you would like to move all rows representing members born before 1960 into a separate partition. As we have already seen, this cannot be done using ALTER TABLE ... ADD PARTITION. However, you can use another partition-related extension to ALTER TABLE in order to accomplish this:

ALTER TABLE members REORGANIZE PARTITION p0 INTO (
    PARTITION s0 VALUES LESS THAN (1960),
    PARTITION s1 VALUES LESS THAN (1970)
);

In effect, this command splits partition p0 into two new partitions s0 and s1. It also moves the data that was stored in p0 into the new partitions according to the rules embodied in the two PARTITION ... VALUES ... clauses, so that s0 contains only those records for which YEAR(dob) is less than 1960 and s1 contains those rows in which YEAR(dob) is greater than or equal to 1960 but less than 1970.

A REORGANIZE PARTITION clause may also be used for merging adjacent partitions. You can return the members table to its previous partitioning as shown here:

ALTER TABLE members REORGANIZE PARTITION s0,s1 INTO (
    PARTITION p0 VALUES LESS THAN (1970)
);

No data is lost in splitting or merging partitions using REORGANIZE PARTITION. In executing the above statement, MySQL moves all of the records that were stored in partitions s0 and s1 into partition p0.

The general syntax for REORGANIZE PARTITION is:

ALTER TABLE tbl_name 
    REORGANIZE PARTITION partition_list 
    INTO (partition_definitions);

Here, tbl_name is the name of the partitioned table, and partition_list is a comma-separated list of names of one or more existing partitions to be changed. partition_definitions is a comma-separated list of new partition definitions, which follow the same rules as for the partition_definitions list used in CREATE TABLE (see Section 13.1.10, “CREATE TABLE Syntax”). It should be noted that you are not limited to merging several partitions into one, or to splitting one partition into many, when using REORGANIZE PARTITION. For example, you can reorganize all four partitions of the members table into two, as follows:

ALTER TABLE members REORGANIZE PARTITION p0,p1,p2,p3 INTO (
    PARTITION m0 VALUES LESS THAN (1980),
    PARTITION m1 VALUES LESS THAN (2000)
);

You can also use REORGANIZE PARTITION with tables that are partitioned by LIST. Let us return to the problem of adding a new partition to the list-partitioned tt table and failing because the new partition had a value that was already present in the value-list of one of the existing partitions. We can handle this by adding a partition that contains only non-conflicting values, and then reorganizing the new partition and the existing one so that the value which was stored in the existing one is now moved to the new one:

ALTER TABLE tt ADD PARTITION (PARTITION np VALUES IN (4, 8));
ALTER TABLE tt REORGANIZE PARTITION p1,np INTO (
    PARTITION p1 VALUES IN (6, 18),
    PARTITION np VALUES in (4, 8, 12)
);

Here are some key points to keep in mind when using ALTER TABLE ... REORGANIZE PARTITION to repartition tables that are partitioned by RANGE or LIST:

Tables which are partitioned by hash or by key are very similar to one another with regard to making changes in a partitioning setup, and both differ in a number of ways from tables which have been partitioned by range or list. For that reason, this section addresses the modification of tables partitioned by hash or by key only. For a discussion of adding and dropping of partitions of tables that are partitioned by range or list, see Section 16.3.1, “Management of RANGE and LIST Partitions”.

You cannot drop partitions from tables that are partitioned by HASH or KEY in the same way that you can from tables that are partitioned by RANGE or LIST. However, you can merge HASH or KEY partitions using the ALTER TABLE ... COALESCE PARTITION command. For example, suppose that you have a table containing data about clients, which is divided into twelve partitions. The clients table is defined as shown here:

CREATE TABLE clients (
    id INT,
    fname VARCHAR(30),
    lname VARCHAR(30),
    signed DATE
)
PARTITION BY HASH( MONTH(signed) )
PARTITIONS 12;

To reduce the number of partitions from twelve to eight, execute the following ALTER TABLE command:

mysql> ALTER TABLE clients COALESCE PARTITION 4;
Query OK, 0 rows affected (0.02 sec)

COALESCE works equally well with tables that are partitioned by HASH, KEY, LINEAR HASH, or LINEAR KEY. Here is an example similar to the previous one, differing only in that the table is partitioned by LINEAR KEY:

mysql> CREATE TABLE clients_lk (
    ->     id INT,
    ->     fname VARCHAR(30),
    ->     lname VARCHAR(30),
    ->     signed DATE
    -> )
    -> PARTITION BY LINEAR KEY(signed)
    -> PARTITIONS 12;
Query OK, 0 rows affected (0.03 sec)

mysql> ALTER TABLE clients_lk COALESCE PARTITION 4;
Query OK, 0 rows affected (0.06 sec)
Records: 0  Duplicates: 0  Warnings: 0

Note that the number following COALESCE PARTITION is the number of partitions to merge into the remainder — in other words, it is the number of partitions to remove from the table.

If you attempt to remove more partitions than the table has, the result is an error like the one shown:

mysql> ALTER TABLE clients COALESCE PARTITION 18;
ERROR 1478 (HY000): Cannot remove all partitions, use DROP TABLE instead

To increase the number of partitions for the clients table from 12 to 18. use ALTER TABLE ... ADD PARTITION as shown here:

ALTER TABLE clients ADD PARTITION PARTITIONS 6;

A number of partitioning maintenance tasks can be carried out in MySQL 5.1. MySQL does not support the commands CHECK TABLE, OPTIMIZE TABLE, ANALYZE TABLE, or REPAIR TABLE for partitioned tables. Instead, you can use a number of extensions to ALTER TABLE which were implemented in MySQL 5.1.5. These can be used for performing operations of this type on one or more partitions directly, as described in the following list:

You can also use the mysqlcheck or myisamchk utility to accomplish these tasks, operating on the separate .MYI files generated by partitioning a table. See Section 8.12, “mysqlcheck — A Table Maintenance and Repair Program”.

This section discusses obtaining information about existing partitions, which can be done in a number of ways. These include:

As discussed elsewhere in this chapter, SHOW CREATE TABLE includes in its output the PARTITION BY clause used to create a partitioned table. For example:

mysql> SHOW CREATE TABLE trb3\G
*************************** 1. row ***************************
       Table: trb3
Create Table: CREATE TABLE `trb3` (
  `id` int(11) default NULL,
  `name` varchar(50) default NULL,
  `purchased` date default NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1 
PARTITION BY RANGE (YEAR(purchased)) (
  PARTITION p0 VALUES LESS THAN (1990) ENGINE = MyISAM, 
  PARTITION p1 VALUES LESS THAN (1995) ENGINE = MyISAM, 
  PARTITION p2 VALUES LESS THAN (2000) ENGINE = MyISAM, 
  PARTITION p3 VALUES LESS THAN (2005) ENGINE = MyISAM
)
1 row in set (0.00 sec)

Note: In early MySQL 5.1 releases, the PARTITIONS clause was not shown for tables partitioned by HASH or KEY. This issue was fixed in MySQL 5.1.6.

SHOW TABLE STATUS works with partitioned tables. Beginning with MySQL 5.1.9, its output is the same as that for non-partitioned tables, except that the Create_options column contains the string partitioned. In MySQL 5.1.8 and earlier, the Engine column always contained the value PARTITION; beginning with MySQL 5.1.9, this column contains the name of the storage engine used by all partitions of the table. (See Section 13.5.4.28, “SHOW TABLE STATUS Syntax”, for more information about this command.)

You can also obtain information about partitions from INFORMATION_SCHEMA, which contains a PARTITIONS table. See Section 22.19, “The INFORMATION_SCHEMA PARTITIONS Table”.

Beginning with MySQL 5.1.5, it is possible to determine which partitions of a partitioned table are involved in a given SELECT query using EXPLAIN PARTITIONS. The PARTITIONS keyword adds a partitions column to the output of EXPLAIN listing the partitions from which records would be matched by the query.

Suppose that you have a table trb1 defined and populated as follows:

CREATE TABLE trb1 (id INT, name VARCHAR(50), purchased DATE)
    PARTITION BY RANGE(id)
    (
        PARTITION p0 VALUES LESS THAN (3),
        PARTITION p1 VALUES LESS THAN (7),
        PARTITION p2 VALUES LESS THAN (9),
        PARTITION p3 VALUES LESS THAN (11)
    );

INSERT INTO trb1 VALUES
    (1, 'desk organiser', '2003-10-15'),
    (2, 'CD player', '1993-11-05'),
    (3, 'TV set', '1996-03-10'),
    (4, 'bookcase', '1982-01-10'),
    (5, 'exercise bike', '2004-05-09'),
    (6, 'sofa', '1987-06-05'),
    (7, 'popcorn maker', '2001-11-22'),
    (8, 'aquarium', '1992-08-04'),
    (9, 'study desk', '1984-09-16'),
    (10, 'lava lamp', '1998-12-25');

You can see which partitions are used in a query such as SELECT * FROM trb1;, as shown here:

mysql> EXPLAIN PARTITIONS SELECT * FROM trb1\G
*************************** 1. row ***************************
           id: 1
  select_type: SIMPLE
        table: trb1
   partitions: p0,p1,p2,p3
         type: ALL
possible_keys: NULL
          key: NULL
      key_len: NULL
          ref: NULL
         rows: 10
        Extra: Using filesort

In this case, all four partitions are searched. However, when a limiting condition making use of the partitioning key is added to the query, you can see that only those partitions containing matching values are scanned, as shown here:

mysql> EXPLAIN PARTITIONS SELECT * FROM trb1 WHERE id < 5\G
*************************** 1. row ***************************
           id: 1
  select_type: SIMPLE
        table: trb1
   partitions: p0,p1
         type: ALL
possible_keys: NULL
          key: NULL
      key_len: NULL
          ref: NULL
         rows: 10
        Extra: Using where

EXPLAIN PARTITIONS provides information about keys used and possible keys, just as with the standard EXPLAIN SELECT statement:

mysql> ALTER TABLE trb1 ADD PRIMARY KEY (id);
Query OK, 10 rows affected (0.03 sec)
Records: 10  Duplicates: 0  Warnings: 0

mysql> EXPLAIN PARTITIONS SELECT * FROM trb1 WHERE id < 5\G
*************************** 1. row ***************************
           id: 1
  select_type: SIMPLE
        table: trb1
   partitions: p0,p1
         type: range
possible_keys: PRIMARY
          key: PRIMARY
      key_len: 4
          ref: NULL
         rows: 7
        Extra: Using where

You should take note of the following restrictions and limitations on EXPLAIN PARTITIONS:

See also Section 7.2.1, “Optimizing Queries with EXPLAIN”.

This section discusses the relationship of partitioning keys with primary keys and unique keys. The rule governing this relationship can be expressed as follows: All columns used in the partitioning expression for a partitioned table must be part of every unique key that the table may have.

In other words, every unique key on the table must use every column in the table's partitioning expression. (This also includes the table's primary key, since it is by definition a unique key. This particular case is discussed later in this section.) For example, each of the following table creation statements is invalid:

CREATE TABLE t1 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    UNIQUE KEY (col1, col2)
)
PARTITION BY HASH(col3)
PARTITIONS 4;

CREATE TABLE t2 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    UNIQUE KEY (col1),
    UNIQUE KEY (col3)
)
PARTITION BY HASH(col1 + col3)
PARTITIONS 4;

In each case, the proposed table would have at least one unique key that does not include all columns used in the partitioning expression.

Each of the following statements is valid, and represents one way in which the corresponding invalid table creation statement could be made to work:

CREATE TABLE t1 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    UNIQUE KEY (col1, col2, col3)
)
PARTITION BY HASH(col3)
PARTITIONS 4;

CREATE TABLE t2 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    UNIQUE KEY (col1, col3)
)
PARTITION BY HASH(col1 + col3)
PARTITIONS 4;

This example shows the error produced in such cases:

mysql> CREATE TABLE t3 (
    ->     col1 INT NOT NULL,
    ->     col2 DATE NOT NULL,
    ->     col3 INT NOT NULL,
    ->     col4 INT NOT NULL,
    ->     UNIQUE KEY (col1, col2),
    ->     UNIQUE KEY (col3)
    -> )
    -> PARTITION BY HASH(col1 + col3)
    -> PARTITIONS 4;
ERROR 1491 (HY000): A PRIMARY KEY must include all columns in the table's partitioning function

The CREATE statement fails because both col1 and col3 are included in the proposed partitioning key, but neither of these columns is part of both of unique keys on the table. This shows one possible fix for the invalid table definition;

mysql> CREATE TABLE t3 (
    ->     col1 INT NOT NULL,
    ->     col2 DATE NOT NULL,
    ->     col3 INT NOT NULL,
    ->     col4 INT NOT NULL,
    ->     UNIQUE KEY (col1, col2, col3),
    ->     UNIQUE KEY (col3)
    -> )
    -> PARTITION BY HASH(col3)
    -> PARTITIONS 4;
Query OK, 0 rows affected (0.05 sec)

In this case, the proposed partitioning key col3 is part of both unique keys, and the table creation statement succeeds.

Since every primary key is by definition a unique key, this restriction also includes the table's primary key, if it has one. For example, the next two statements are invalid:

CREATE TABLE t4 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    PRIMARY KEY(col1, col2)
)
PARTITION BY HASH(col3)
PARTITIONS 4;

CREATE TABLE t5 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    PRIMARY KEY(col1, col3),
    UNIQUE KEY(col2)
)
PARTITION BY HASH( YEAR(col2) )
PARTITIONS 4;

In both cases, the primary key does not include all columns referenced in the partitioning expression. However, both of the next two statements are valid:

CREATE TABLE t6 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    PRIMARY KEY(col1, col2)
)
PARTITION BY HASH(col1 + YEAR(col2))
PARTITIONS 4;

CREATE TABLE t7 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    PRIMARY KEY(col1, col2, col4),
    UNIQUE KEY(col2, col1)
)
PARTITION BY HASH(col1 + YEAR(col2))
PARTITIONS 4;

If a table has no unique keys — this includes having no primary key — then this restriction does not apply, and you may use any column or columns in the partitioning expression as long as the column type is compatible with the partitioning type.

For the same reason, you cannot later add a unique key to a partitioned table unless the key includes all columns used by the table's partitioning expression. Consider given the partitioned table defined as shown here:

CREATE TABLE t_no_pk (c1 INT, c2 INT) 
    PARTITION BY RANGE(c1) (
        PARTITION p0 VALUES LESS THAN (10), 
        PARTITION p1 VALUES LESS THAN (20), 
        PARTITION p2 VALUES LESS THAN (30), 
        PARTITION p3 VALUES LESS THAN (40)
    );

It is possible to add a primary key to t_no_pk using either of these ALTER TABLE statements:

#  possible PK
ALTER TABLE t_no_pk ADD PRIMARY KEY(c1);

#  also a possible PK
ALTER TABLE t_no_pk ADD PRIMARY KEY(c1, c2);

However, the next statement fails, because c1 is part of the partitioning key, but is not part of the proposed primary key:

#  fails with ERROR 1482
ALTER TABLE t_no_pk ADD PRIMARY KEY(c2);

Since t_no_pk has only c1 in its partitioning expression, attempting to adding a unique key on c2 alone fails. However, you can add a unique key that uses both c1 and c2.

These rules also apply to existing non-partitioned tables that you wish to partition using ALTER TABLE ... PARTITION BY. Consider a table np_pk defined as shown here:

CREATE TABLE np_pk (
    id INT NOT NULL AUTO_INCREMENT,
    name VARCHAR(50),
    added DATE,
    PRIMARY KEY (id)
);

The following ALTER TABLE statements fails with an error, because the added column is not part of any unique key in the table:

ALTER TABLE np_pk
    PARTITION BY HASH( TO_DAYS(added) )
    PARTITIONS 4;

This statement, however, would be valid:

ALTER TABLE np_pk
    PARTITION BY HASH(id)
    PARTITIONS 4;

In the case of np_pk, the only column that may be used as part of a partitioning expression is id; if you wish to partition this table using any other column or columns in the partitioning expression, you must first modify the table, either by adding the desired column or columns to the primary key, or by dropping the primary key altogether.

We are working to remove this limitation in a future MySQL release series.

The following limitations apply to the use of storage engines with user-defined partitioning of tables.

MERGE storage engine.  Tables using the MERGE storage engine cannot be partitioned.

FEDERATED storage engine.  Partitioning of FEDERATED tables is not supported. Beginning with MySQL 5.1.15, it is not possible to create partitioned FEDERATED tables at all. We are working to remove this limitation in a future MySQL release.

CSV storage engine.  Partitioned tables using the CSV storage engine are not supported. Starting with MySQL 5.1.12, it is not possible to create partitioned CSV tables at all.

BLACKHOLE storage engine.  Prior to MySQL 5.1.6, tables using the BLACKHOLE storage engine also could not be partitioned.

NDB storage engine (MySQL Cluster).  Partitioning by KEY (or LINEAR KEY) is the only type of partitioning supported for the NDB storage engine. Beginning with MySQL 5.1.12, it is not possible to create a Cluster table using any partitioning type other than [LINEAR] KEY, and attempting to do so gives rise to an error.

Upgrading partitioned tables.  When performing an upgrade, tables which are partitioned by KEY and that use any storage engine other than NDBCLUSTER and must be dumped and reloaded.

Same storage engine for all partitions.  All of a table's partitions and subpartitions (if there are any of the latter) must use the same storage engine. We are working to remove this limitation in a future MySQL release.

This section discusses limitations in MySQL Partitioning relating specifically to functions used in partitioning expressions, including both those function which are specifically supported for use in partitioning expressions, and those which are specifically prohibited.