SQL GROUP BY
Summary: in this tutorial, you will learn how to use the SQL GROUP BY
clause to group rows based on one or more columns.
Mục Lục
Introduction to SQL GROUP BY clause
The GROUP BY
is an optional clause of the SELECT
statement. The GROUP BY
clause allows you to group rows based on values of one or more columns. It returns one row for each group.
The following shows the basic syntax of the GROUP BY
clause:
SELECT
column1, column2, aggregate_function(column3)FROM
table_nameGROUP
BY
column1, column2;Code language:
SQL (Structured Query Language)
(
sql
)
The following picture illustrates shows how the GROUP BY
clause works:
The table on the left side has two columns id
and fruit
. When you apply the GROUP BY
clause to the fruit
column, it returns the result set that includes unique values from the fruit
column:
SELECT fruit FROM sample_table GROUP BY fruit;
In practice, you often use the GROUP BY
clause with an aggregate function such as MIN, MAX, AVG, SUM, or COUNT to calculate a measure that provides the information for each group.
For example, the following illustrates how the GROUP BY
clause works with the COUNT
aggregate function:
In this example, we group the rows by the values of the fruit
column and apply the COUNT
function to the id
column. The result set includes the unique values of the fruit columns and the number of the corresponding rows.
SELECT fruit, COUNT(id) FROM sample_table GROUP BY fruit;
The columns that appear in the GROUP BY
clause are called grouping columns. If a grouping column contains NULL values, all NULL values are summarized into a single group because the GROUP BY
clause considers all NULL values equal.
SQL GROUP BY examples
We will use the employees
and departments
tables in the sample database to demonstrate how the GROUP BY
clause works.
The following example uses the GROUP BY
clause to group the values in department_id
column of the employees
table:
SELECT
department_idFROM
employeesGROUP
BY
department_id;Code language:
SQL (Structured Query Language)
(
sql
)
Output:
Try It
+---------------+ | department_id | +---------------+ |
1
| |2
| |3
| |4
| |5
| |6
| |7
| |8
| |9
| |10
| |11
| +---------------+11
rowsin
set
(0.00 sec)Code language:
JavaScript
(
javascript
)
In this example:
- First, the
SELECT
clause returns all values from the department_id column ofemployees
table. - Second, the
GROUP BY
clause groups all values into groups.
The department_id
column of the employees
table has 40 rows, including duplicate department_id
values. However, the GROUP BY
groups these values into groups.
Without an aggregate function, the GROUP BY
behaves like the DISTINCT
keyword:
SELECT
DISTINCT
department_idFROM
employeesORDER
BY
department_id;Code language:
SQL (Structured Query Language)
(
sql
)
Try It
The GROUP BY
clause will be more useful when you use it with an aggregate function.
For example, the following statement uses the GROUP BY
clause with the COUNT
function to count the number of employees by department:
SELECT
department_id,COUNT
(employee_id) headcountFROM
employeesGROUP
BY
department_id;Code language:
SQL (Structured Query Language)
(
sql
)
Try It
Output:
+---------------+-----------+ | department_id | headcount | +---------------+-----------+ |
1
|1
| |2
|2
| |3
|6
| |4
|1
| |5
|7
| |6
|5
| |7
|1
| |8
|6
| |9
|3
| |10
|6
| |11
|2
| +---------------+-----------+11
rowsin
set
(0.00 sec)Code language:
JavaScript
(
javascript
)
How it works.
- First, the
GROUP BY
clause groups the rows in theemployees
table by department id. - Second, the
COUNT(employee_id)
returns the number of employee id values in each group.
SQL GROUP BY with INNER JOIN example
The following example returns the number of employees by department. And it uses an INNER JOIN
clause to include the department name in the result:
SELECT
department_name,COUNT
(employee_id) headcountFROM
employees eINNER
JOIN
departments dON
d.department_id = e.department_idGROUP
BY
department_name;Code language:
SQL (Structured Query Language)
(
sql
)
Try It
Output:
+------------------+-----------+ | department_name | headcount | +------------------+-----------+ | Accounting |
2
| | Administration |1
| | Executive |3
| | Finance |6
| | Human Resources |1
| | IT |5
| | Marketing |2
| | Public Relations |1
| | Purchasing |6
| | Sales |6
| | Shipping |7
| +------------------+-----------+11
rowsin
set
(0.01 sec)Code language:
JavaScript
(
javascript
)
SQL GROUP BY with ORDER BY example
The following example uses an ORDER BY clause to sort the departments by headcount:
SELECT
department_name,COUNT
(employee_id) headcountFROM
employees eINNER
JOIN
departments dON
d.department_id = e.department_idGROUP
BY
department_nameORDER
BY
headcountDESC
;Code language:
SQL (Structured Query Language)
(
sql
)
Try It
Output:
+------------------+-----------+ | department_name | headcount | +------------------+-----------+ | Shipping |
7
| | Sales |6
| | Finance |6
| | Purchasing |6
| | IT |5
| | Executive |3
| | Marketing |2
| | Accounting |2
| | Human Resources |1
| | Administration |1
| | Public Relations |1
| +------------------+-----------+11
rowsin
set
(0.00 sec)Code language:
JavaScript
(
javascript
)
Note that you can use either the headcount
alias or the COUNT(employee_id)
in the ORDER BY
clause.
SQL GROUP BY with HAVING example
The following example uses the HAVING
clause to find departments with headcounts are greater than 5:
SELECT
department_name,COUNT
(employee_id) headcountFROM
employees eINNER
JOIN
departments dON
d.department_id = e.department_idGROUP
BY
department_nameHAVING
headcount >5
ORDER
BY
headcountDESC
;Code language:
SQL (Structured Query Language)
(
sql
)
Try It
Output:
+-----------------+-----------+ | department_name | headcount | +-----------------+-----------+ | Shipping |
7
| | Sales |6
| | Finance |6
| | Purchasing |6
| +-----------------+-----------+4
rowsin
set
(0.00 sec)Code language:
JavaScript
(
javascript
)
SQL GROUP BY with MIN, MAX, and AVG example
The following query returns the minimum, maximum, and average salary of employees in each department.
SELECT
department_name,MIN
(salary) min_salary,MAX
(salary) max_salary,ROUND
(AVG
(salary),2
) average_salaryFROM
employees eINNER
JOIN
departments dON
d.department_id = e.department_idGROUP
BY
department_name;Code language:
SQL (Structured Query Language)
(
sql
)
Try It
Output:
+------------------+------------+------------+----------------+ | department_name | min_salary | max_salary | average_salary | +------------------+------------+------------+----------------+ | Accounting |
8300.00
|12000.00
|10150.00
| | Administration |4400.00
|4400.00
|4400.00
| | Executive |17000.00
|24000.00
|19333.33
| | Finance |6900.00
|12000.00
|8600.00
| | Human Resources |6500.00
|6500.00
|6500.00
| | IT |4200.00
|9000.00
|5760.00
| | Marketing |6000.00
|13000.00
|9500.00
| | Public Relations |10000.00
|10000.00
|10000.00
| | Purchasing |2500.00
|11000.00
|4150.00
| | Sales |6200.00
|14000.00
|9616.67
| | Shipping |2700.00
|8200.00
|5885.71
| +------------------+------------+------------+----------------+11
rowsin
set
(0.01 sec)Code language:
JavaScript
(
javascript
)
SQL GROUP BY with SUM function example
To get the total salary per department, you apply the SUM function to the salary
column and group employees by the department_id
column as follows:
SELECT
department_name,SUM
(salary) total_salaryFROM
employees eINNER
JOIN
departments dON
d.department_id = e.department_idGROUP
BY
department_name;Code language:
SQL (Structured Query Language)
(
sql
)
Try It
Output:
+------------------+--------------+ | department_name | total_salary | +------------------+--------------+ | Accounting |
20300.00
| | Administration |4400.00
| | Executive |58000.00
| | Finance |51600.00
| | Human Resources |6500.00
| | IT |28800.00
| | Marketing |19000.00
| | Public Relations |10000.00
| | Purchasing |24900.00
| | Sales |57700.00
| | Shipping |41200.00
| +------------------+--------------+11
rowsin
set
(0.01 sec)Code language:
JavaScript
(
javascript
)
SQL GROUP BY multiple columns
So far, you have seen that we have grouped all employees by one column. For example, the following clause places all rows with the same values in the department_id
column in one group.
GROUP BY department_id
Code language:
SQL (Structured Query Language)
(
sql
)
How about grouping employees by values in both department_id
and job_id
columns?
GROUP BY department_id, job_id
Code language:
SQL (Structured Query Language)
(
sql
)
This clause will group all employees with the same values in both department_id
and job_id
columns in one group.
The following statement groups rows with the same values in both department_id
and job_id
columns in the same group then return the rows for each of these groups.
SELECT
department_name, job_title,COUNT
(employee_id)FROM
employees eINNER
JOIN
departments dON
d.department_id = e.department_idINNER
JOIN
jobs jON
j.job_id = e.job_idGROUP
BY
department_name , job_title;Code language:
SQL (Structured Query Language)
(
sql
)
Try It
+------------------+---------------------------------+--------------------+ | department_name | job_title | COUNT(employee_id) | +------------------+---------------------------------+--------------------+ | Accounting | Accounting Manager |
1
| | Accounting |Public
Accountant |1
| | Administration | Administration Assistant |1
| | Executive | Administration Vice President |2
| | Executive | President |1
| | Finance | Accountant |5
| | Finance | Finance Manager |1
| | Human Resources | Human Resources Representative |1
| | IT | Programmer |5
| | Marketing | Marketing Manager |1
| | Marketing | Marketing Representative |1
| |Public
Relations |Public
Relations Representative |1
| | Purchasing | Purchasing Clerk |5
| | Purchasing | Purchasing Manager |1
| | Sales | Sales Manager |2
| | Sales | Sales Representative |4
| | Shipping | Shipping Clerk |2
| | Shipping | Stock Clerk |1
| | Shipping | Stock Manager |4
| +------------------+---------------------------------+--------------------+19
rows in set (0.00
sec)Code language:
PHP
(
php
)
Summary
- The
GROUP BY
clause groups the rows into groups based on the values of one or more columns. - Use an aggregate function with the
GROUP BY
clause to calculate the summarized value for each group.
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