The MySQL Partitioning is a technique that can be used to divide a database table into smaller tables i.e. partitions. These smaller tables are stored in different physical locations and are treated as separate tables. Thus, the data in these smaller tables can be accessed and managed individually.

But note that, even if the data smaller tables is managed separately, they are not independent tables; i.e., they are still a part of main table.

There are two forms of partitioning in MySQL: Horizontal Partitioning and Vertical Partitioning.

MySQL Horizontal Partitioning

The MySQL Horizontal partitioning is used to divide the table rows into multiple partitions. Since it divides the rows, all the columns will be present in each partition. All the partitions can be accessed individually or collectively.

There are several types of MySQL horizontal partitioning methods −

MySQL Range Partitioning

The MySQL RANGE partitioning is used to divide a table into partitions based on a specific range of column values. Each table partition contains rows with column values falling within that defined range.

Example

Let us create a table named CUSTOMERS and partition it by the AGE column into four partitions: P1, P2, P3, and P4 using the “PARTITION BY RANGE” clause −

CREATE TABLE CUSTOMERS(
   ID int not null,
   NAME varchar(40) not null,
   AGE int not null,
   ADDRESS char(25) not null,
   SALARY decimal(18, 2)
   )
   PARTITION BY RANGE (AGE) (
   PARTITION P1 VALUES LESS THAN (20),
   PARTITION P2 VALUES LESS THAN (30),
   PARTITION P3 VALUES LESS THAN (40),
   PARTITION P4 VALUES LESS THAN (50)
);

Here, we are inserting rows into the above created table −

INSERT INTO CUSTOMERS VALUES 
(1, ''Ramesh'', 19, ''Ahmedabad'', 2000.00 ),
(2, ''Khilan'', 25, ''Delhi'', 1500.00 ),
(3, ''kaushik'', 23, ''Kota'', 2000.00 ),
(4, ''Chaitali'', 31, ''Mumbai'', 6500.00 ),
(5, ''Hardik'', 35, ''Bhopal'', 8500.00 ),
(6, ''Komal'', 47, ''MP'', 4500.00 ),
(7, ''Muffy'', 43, ''Indore'', 10000.00 );

Following is the CUSTOMERS table obtained −

ID	NAME	AGE	ADDRESS	SALARY
1	Ramesh	19	Ahmedabad	2000.00
2	Khilan	25	Delhi	1500.00
3	Kaushik	23	Kota	2000.00
4	Chaitali	31	Mumbai	6500.00
5	Hardik	35	Bhopal	8500.00
6	Komal	47	MP	4500.00
7	Muffy	43	Indore	10000.00

Now that we have some data in the CUSTOMERS table, we can display the partition status to see how the data is distributed among the partitions using the following query −

SELECT PARTITION_NAME, TABLE_ROWS 
FROM INFORMATION_SCHEMA.PARTITIONS 
WHERE TABLE_NAME=''CUSTOMERS

The above query will show us the number of rows in each partition. For example, P1 has 1 row, P2 has 2 rows, P3 has 2 rows, and P4 has 2 rows as shown below −

PARTITION_NAME	TABLE_ROWS
P1	1
P2	2
P3	2
P4	2

Displaying Partitions −

We can also display data from specific partitions using the PARTITION clause. For instance, to retrieve data from partition P1, we use the following query −

SELECT * FROM CUSTOMERS PARTITION (p1);

It will display all the records in partition P1 −

ID	NAME	AGE	ADDRESS	SALARY
1	Ramesh	19	Ahmedabad	2000.00

Similarly, we can display other partitions using the same syntax.

Handling Data Outside the Range −

If we attempt to insert a value into the AGE column that doesn”t fall within any of the defined partitions, it will fail with an error, as shown below −

INSERT INTO CUSTOMERS VALUES 
(8, ''Brahmi'', 70, ''Hyderabad'', 19000.00 );

Following is the error obtained −

ERROR 1526 (HY000): Table has no partition for value 70

Truncating Partitions −

We can also manage partitions by truncating them if needed. For example, to empty partition P2, we can use the following query −

ALTER TABLE CUSTOMERS TRUNCATE PARTITION p2;

The output obtained is as shown below −

Query OK, 0 rows affected (0.03 sec)

This will remove all data from partition P2, making it empty as shown below −

SELECT * FROM CUSTOMERS PARTITION (p2);

Following is the output produced −

Empty set (0.00 sec)

We can verify the CUSTOMERS table using the following SELECT query −

SELECT * FROM CUSTOMERS;

We can see in the table below that the rows belonging to p2 partition are deleted −

ID	NAME	AGE	ADDRESS	SALARY
1	Ramesh	19	Ahmedabad	2000.00
2	Khilan	25	Delhi	1500.00
3	Kaushik	23	Kota	2000.00
6	Komal	47	MP	4500.00
7	Muffy	43	Indore	10000.00

MySQL List Partitioning

The MySQL List Partitioning is used to divide the table into partitions based on a discrete set of values for a specific column. Each partition contains rows that match a particular value within the defined set.

Example

In this example, we will create a table named STUDENTS and divide it into four partitions (P1, P2, P3, and P4) based on the “DEPARTMENT_ID” column using the “PARTITION BY LIST” clause −

CREATE TABLE STUDENTS(
   ID int,
   NAME varchar(50),
   DEPARTMENT varchar(50),
   DEPARTMENT_ID int
   )
   PARTITION BY LIST(DEPARTMENT_ID)(
   PARTITION P1 VALUES IN (3, 5, 6, 7, 9),
   PARTITION P2 VALUES IN (13, 15, 16, 17, 20),
   PARTITION P3 VALUES IN (23, 25, 26, 27, 30),
   PARTITION P4 VALUES IN (33, 35, 36, 37, 40)
);

Here, we are inserting rows into the above-created table −

INSERT INTO STUDENTS VALUES 
(1, ''Ramesh'', "cse", 5),
(2, ''Khilan'', "mech", 20),
(3, ''kaushik'', "ece", 17),
(4, ''Chaitali'', "eee", 33),
(5, ''Hardik'', "IT", 36),
(6, ''Komal'', "Hotel management", 40),
(7, ''Muffy'', "Fashion", 23);

Following is the STUDENTS table obtained −

ID	NAME	DEPARTMENT	DEPARTMENT_ID
1	Ramesh	cse	5
2	Khilan	mech	20
3	Kaushik	ece	17
7	Muffy	Fashion	23
4	Chaitali	eee	33
5	Hardik	IT	36
6	Komal	Hotel management	40

We can display the partition status of the STUDENTS table to see how the data is distributed among partitions using the following query −

SELECT PARTITION_NAME, TABLE_ROWS 
FROM INFORMATION_SCHEMA.PARTITIONS 
WHERE TABLE_NAME=''STUDENTS

The output of this query will show the number of rows in each partition. For instance, P1 has 1 row, P2 has 2 rows, P3 has 1 row, and P4 has 3 rows −

PARTITION_NAME	TABLE_ROWS
P1	1
P2	2
P3	1
P4	3

MySQL Hash Partitioning

The MySQL HASH partitioning is used to divide the table data into partitions using a hash function based on a specific column(s). The data will be evenly distributed among the partitions.

Example

In the following query, we are creating a table with the name EMPLOYEES with four partitions based on the “id” column using the PARTITION BY HASH clause −

CREATE TABLE EMPLOYEES (
   id INT NOT NULL,
   name VARCHAR(50) NOT NULL,
   department VARCHAR(50) NOT NULL,
   salary INT NOT NULL
 )
   PARTITION BY HASH(id)
   PARTITIONS 4;

Here, we are inserting rows into the above-created table −

INSERT INTO EMPLOYEES VALUES 
(1, ''Varun'', ''Sales'', 50000),
(2, ''Aarohi'', ''Marketing'', 60000),
(3, ''Paul'', ''IT'', 70000),
(4, ''Vaidhya'', ''Finance'', 80000),
(5, ''Nikhil'', ''Sales'', 55000),
(6, ''Sarah'', ''Marketing'', 65000),
(7, ''Tim'', ''IT'', 75000),
(8, ''Priya'', ''Finance'', 85000);

The EMPLOYEES table obtained is as follows −

id	name	department	salary
4	Vaidhya	Finance	80000
8	Priya	Finance	85000
1	Varun	Sales	50000
5	Nikhil	Sales	55000
2	Aarohi	Marketing	60000
6	Sarah	Marketing	65000
3	Paul	IT	70000
7	Tim	IT	75000

The records are evenly distributed among four partitions based on the “id” column. You can verify the partition status using the following SELECT query −

SELECT PARTITION_NAME, TABLE_ROWS 
FROM INFORMATION_SCHEMA.PARTITIONS 
WHERE TABLE_NAME=''EMPLOYEES

The table obtained is as follows −

PARTITION_NAME	TABLE_ROWS
P0	2
P1	2
P2	2
P3	2

Key Partitioning

The MySQL key partitioning is used to divide the table data into partitions based on the values of the primary key or a unique key.

Example

In the following query, we are creating a table with the name PERSON with Key partitioning on the “id” column. We have divided the table into four partitions, and the primary key is “id” −

CREATE TABLE PERSON (
   id INT NOT NULL,
   name VARCHAR(50) NOT NULL,
   email VARCHAR(50) NOT NULL,
   address VARCHAR(100) NOT NULL,
   PRIMARY KEY (id)
   )
   PARTITION BY KEY(id)
   PARTITIONS 4;

Here, we are inserting rows into the above-created table −

INSERT INTO PERSON VALUES 
(1, ''Krishna'', ''Krishna@tutorialspoint.com'', ''Ayodhya''),
(2, ''Kasyap'', ''Kasyap@tutorialspoint.com'', ''Ayodhya''),
(3, ''Radha'', ''Radha@tutorialspoint.com'', ''Ayodhya''),
(4, ''Sarah'', ''Sarah@tutorialspoint.com'', ''Sri Lanka''),
(5, ''Sita'', ''Sita@tutorialspoint.com'', ''Sri Lanka''),
(6, ''Arjun'', ''Arjun@tutorialspoint.com'', ''India''),
(7, ''Hanuman'', ''Hanuman@tutorialspoint.com'', ''Sri Lanka''),
(8, ''Lakshman'', ''Lakshman@tutorialspoint.com'', ''Sri Lanka'');

Following is the PERSON table obtained −

id	name	email	address
1	Krishna	Krishna@tutorialspoint.com	Ayodhya
5	Sita	Sita@tutorialspoint.com	Sri Lanka
4	Sarah	Sarah@tutorialspoint.com	Sri Lanka
8	Lakshman	Lakshman@tutorialspoint.com	Sri Lanka
3	Radha	Radha@tutorialspoint.com	Ayodhya
7	Hanuman	Hanuman@tutorialspoint.com	Sri Lanka
2	Kasyap	Kasyap@tutorialspoint.com	Ayodhya
6	Arjun	Arjun@tutorialspoint.com	India

Again, the data is evenly distributed among partitions based on the “id” column, and you can verify the partition status using the query given below −

SELECT PARTITION_NAME, TABLE_ROWS 
FROM INFORMATION_SCHEMA.PARTITIONS 
WHERE TABLE_NAME=''PERSON

The output obtained is as shown below −

PARTITION_NAME	TABLE_ROWS
P0	2
P1	2
P2	2
P3	2

MySQL Sub-partitioning

The MySQL subpartitioning is used to further divide partitions based on another column, often used in conjunction with other partitioning methods like RANGE or HASH.

Example

Let us create a CUSTOMER_ORDERS table with RANGE partitioning on the “order_date” column, and then we will subpartition by hashing on the month of “order_date” −

CREATE TABLE CUSTOMER_ORDERS (
   order_id INT NOT NULL,
   customer_name VARCHAR(50) NOT NULL,
   order_date DATE NOT NULL,
   order_status VARCHAR(20) NOT NULL
   )
   PARTITION BY RANGE (YEAR(order_date))
   SUBPARTITION BY HASH(MONTH(order_date))
   SUBPARTITIONS 2(
   PARTITION p0 VALUES LESS THAN (2022),
   PARTITION p1 VALUES LESS THAN (2023),
   PARTITION p2 VALUES LESS THAN (2024)
);

Here, we are inserting rows into the above-created table −

INSERT INTO CUSTOMER_ORDERS VALUES 
(1, ''John'', ''2021-03-15'', ''Shipped''),
(2, ''Bob'', ''2019-01-10'', ''Delivered''),
(3, ''Johnson'', ''2023-01-10'', ''Delivered''),
(4, ''Jake'', ''2020-01-10'', ''Delivered''),
(5, ''Smith'', ''2022-05-01'', ''Pending''),
(6, ''Rob'', ''2023-01-10'', ''Delivered'');

Following is the CUSTOMERS_ORDERS table obtained −

order_id	customer_name	order_date	order_status
1	John	2021-03-15	Shipped
2	Bob	2019-01-10	Delivered
4	Jake	2020-01-10	Delivered
5	Smith	2022-05-01	Pending
3	Johnson	2023-01-10	Delivered
6	Rob	2023-01-10	Delivered

You can display the CUSTOMER_ORDERS table and verify the partition status using the following query −

SELECT PARTITION_NAME, TABLE_ROWS 
FROM INFORMATION_SCHEMA.PARTITIONS 
WHERE TABLE_NAME=''CUSTOMER_ORDERS

Following is the table obtained −

PARTITION_NAME	TABLE_ROWS
P0	0
P0	3
P1	0
P1	1
P2	0
P2	2