SQL GROUP BY clause
GROUP BY clause
The usage of SQL GROUP BY clause is, to divide the rows in a table into smaller groups.
The GROUP BY clause is used with the SQL SELECT statement.
The grouping can happen after retrieves the rows from a table.
When some rows are retrieved from a grouped result against some condition, that is possible with HAVING clause.
The GROUP BY clause is used with the SELECT statement to make a group of rows based on the values of a specific column or expression. The SQL AGGREGATE function can be used to get summary information for every group and these are applied to an individual group.
The WHERE clause is used to retrieve rows based on a certain condition, but it can not be applied to grouped result.
In an SQL statement, suppose you are using GROUP BY, if required you can use HAVING instead of WHERE, after GROUP BY.
Syntax:
SELECT <column_list> FROM < table name > WHERE <condition>GROUP BY <columns> [HAVING] <condition>;
Parameters:
| Name | Description |
|---|---|
| table_name | Name of the table. |
| column_list | Name of the columns of the table. |
| columns | Name of the columns which will participate in grouping.. |
Pictorial Presentation of Groups of Data
Some important questions related to the SQL GROUP BY clause:
What is the purpose of the SQL GROUP BY clause?
Can you use multiple columns in the GROUP BY clause?
What is the difference between WHERE and HAVING clauses when used with GROUP BY?
Can you sort the result set using columns that are not included in the GROUP BY clause?
What happens if you use a column in the SELECT list that is not included in the GROUP BY clause?
Can you use aliases in the GROUP BY clause?
What is the order of execution for clauses in a SQL query with GROUP BY?
How do you include aggregated values in the result set along with grouped columns?
Can you use aggregate functions without the GROUP BY clause?
What are some common mistakes to avoid when using the GROUP BY clause?
Using GROUP BY with Aggregate Functions
- The power of aggregate functions is greater when combined with the GROUP BY clause.
- The GROUP BY clause is rarely used without an aggregate function.
SQL GROUP BY with COUNT() function
The following query displays number of employees work in each department.
Sample table: employees
employee_id first_name last_name email phone_number hire_date job_id salary commission_pct manager_id department_id ----------- ---------- ---------- ---------- ------------ ---------- ---------- ---------- -------------- ---------- ------------- 100 Steven King SKING 515.123.4567 6/17/1987 AD_PRES 24000 90 101 Neena Kochhar NKOCHHAR 515.123.4568 6/18/1987 AD_VP 17000 100 90 102 Lex De Haan LDEHAAN 515.123.4569 6/19/1987 AD_VP 17000 100 90 103 Alexander Hunold AHUNOLD 590.423.4567 6/20/1987 IT_PROG 9000 102 60 104 Bruce Ernst BERNST 590.423.4568 6/21/1987 IT_PROG 6000 103 60 105 David Austin DAUSTIN 590.423.4569 6/22/1987 IT_PROG 4800 103 60 106 Valli Pataballa VPATABAL 590.423.4560 6/23/1987 IT_PROG 4800 103 60 107 Diana Lorentz DLORENTZ 590.423.5567 6/24/1987 IT_PROG 4200 103 60 108 Nancy Greenberg NGREENBE 515.124.4569 6/25/1987 FI_MGR 12000 101 100 109 Daniel Faviet DFAVIET 515.124.4169 6/26/1987 FI_ACCOUNT 9000 108 100 ........... 206 William Gietz WGIETZ 515.123.8181 10/1/1987 AC_ACCOUNT 8300 205 110
SQL Code:
-- This SQL query calculates the number of employees in each department and displays the department code along with the count.
-- SELECT statement begins
SELECT
department_id "Department Code", -- Selects the 'department_id' column and renames it as 'Department Code' in the result set
COUNT(*) "No of Employees" -- Counts the number of rows for each department and renames the result column as 'No of Employees'
FROM
employees -- Specifies the 'employees' table to retrieve data from
GROUP BY
department_id; -- Groups the result set by the 'department_id' column
Explanation:
- This SQL code is a SELECT statement that retrieves data from the 'employees' table.
- It calculates the number of employees in each department and displays the department code along with the count.
- The SELECT clause selects the 'department_id' column from the 'employees' table and renames it as 'Department Code' in the result set.
- The COUNT(*) function counts the number of rows for each department.
- The result column from the COUNT(*) function is renamed as 'No of Employees'.
- The GROUP BY clause groups the result set by the 'department_id' column, ensuring that the count is calculated for each unique department.
- The query does not filter or manipulate the data further; it simply retrieves the department code and the count of employees for each department from the 'employees' table.
Output:
Department Code No of Employees
--------------- ---------------
100 6
30 6
1
90 3
20 2
70 1
110 2
50 45
80 34
40 1
60 5
10 1
SQL GROUP BY with SUM() function
The following query displays total salary paid to employees work in each department.
Sample table: employees
employee_id first_name last_name email phone_number hire_date job_id salary commission_pct manager_id department_id ----------- ---------- ---------- ---------- ------------ ---------- ---------- ---------- -------------- ---------- ------------- 100 Steven King SKING 515.123.4567 6/17/1987 AD_PRES 24000 90 101 Neena Kochhar NKOCHHAR 515.123.4568 6/18/1987 AD_VP 17000 100 90 102 Lex De Haan LDEHAAN 515.123.4569 6/19/1987 AD_VP 17000 100 90 103 Alexander Hunold AHUNOLD 590.423.4567 6/20/1987 IT_PROG 9000 102 60 104 Bruce Ernst BERNST 590.423.4568 6/21/1987 IT_PROG 6000 103 60 105 David Austin DAUSTIN 590.423.4569 6/22/1987 IT_PROG 4800 103 60 106 Valli Pataballa VPATABAL 590.423.4560 6/23/1987 IT_PROG 4800 103 60 107 Diana Lorentz DLORENTZ 590.423.5567 6/24/1987 IT_PROG 4200 103 60 108 Nancy Greenberg NGREENBE 515.124.4569 6/25/1987 FI_MGR 12000 101 100 109 Daniel Faviet DFAVIET 515.124.4169 6/26/1987 FI_ACCOUNT 9000 108 100 ........... 206 William Gietz WGIETZ 515.123.8181 10/1/1987 AC_ACCOUNT 8300 205 110
SQL Code:
-- This SQL query calculates the total salary for each department and displays the department ID along with the sum of salaries.
-- SELECT statement begins
SELECT
department_id, -- Selects the 'department_id' column
SUM(salary) -- Calculates the sum of salaries for each department
FROM
employees -- Specifies the 'employees' table to retrieve data from
GROUP BY
department_id; -- Groups the result set by the 'department_id' column
Explanation:
- This SQL code is a SELECT statement that retrieves data from the 'employees' table.
- It calculates the total salary for each department and displays the department ID along with the sum of salaries.
- The SELECT clause selects the 'department_id' column from the 'employees' table.
- The SUM(salary) function calculates the sum of salaries for each department.
- The GROUP BY clause groups the result set by the 'department_id' column, ensuring that the sum is calculated for each unique department.
- The query does not filter or manipulate the data further; it simply retrieves the department ID and the sum of salaries for each department from the 'employees' table.
Relational Algebra Expression:
Relational Algebra Tree:
Output:
DEPARTMENT_ID SUM(SALARY)
------------- -----------
100 51608
30 24900
7000
90 58000
20 19000
70 10000
110 20308
50 156400
80 304500
40 6500
60 28800
10 4400
SQL GROUP BY with COUNT() and SUM() function
The following query displays number of employees, total salary paid to employees work in each department.
Sample table: employees
employee_id first_name last_name email phone_number hire_date job_id salary commission_pct manager_id department_id ----------- ---------- ---------- ---------- ------------ ---------- ---------- ---------- -------------- ---------- ------------- 100 Steven King SKING 515.123.4567 6/17/1987 AD_PRES 24000 90 101 Neena Kochhar NKOCHHAR 515.123.4568 6/18/1987 AD_VP 17000 100 90 102 Lex De Haan LDEHAAN 515.123.4569 6/19/1987 AD_VP 17000 100 90 103 Alexander Hunold AHUNOLD 590.423.4567 6/20/1987 IT_PROG 9000 102 60 104 Bruce Ernst BERNST 590.423.4568 6/21/1987 IT_PROG 6000 103 60 105 David Austin DAUSTIN 590.423.4569 6/22/1987 IT_PROG 4800 103 60 106 Valli Pataballa VPATABAL 590.423.4560 6/23/1987 IT_PROG 4800 103 60 107 Diana Lorentz DLORENTZ 590.423.5567 6/24/1987 IT_PROG 4200 103 60 108 Nancy Greenberg NGREENBE 515.124.4569 6/25/1987 FI_MGR 12000 101 100 109 Daniel Faviet DFAVIET 515.124.4169 6/26/1987 FI_ACCOUNT 9000 108 100 ........... 206 William Gietz WGIETZ 515.123.8181 10/1/1987 AC_ACCOUNT 8300 205 110
SQL Code:
-- This SQL query calculates the number of employees, total salary, and displays the department code for each department.
-- SELECT statement begins
SELECT
department_id "Department Code", -- Selects the 'department_id' column and renames it as 'Department Code' in the result set
COUNT(*) "No of Employees", -- Counts the number of rows for each department and renames the result column as 'No of Employees'
SUM(salary) "Total Salary" -- Calculates the sum of salaries for each department and renames the result column as 'Total Salary'
FROM
employees -- Specifies the 'employees' table to retrieve data from
GROUP BY
department_id; -- Groups the result set by the 'department_id' column
Explanation:
- This SQL code is a SELECT statement that retrieves data from the 'employees' table.
- It calculates the number of employees, total salary, and displays the department code for each department.
- The SELECT clause selects the 'department_id' column from the 'employees' table and renames it as 'Department Code' in the result set.
- The COUNT(*) function counts the number of rows for each department and renames the result column as 'No of Employees'.
- The SUM(salary) function calculates the sum of salaries for each department and renames the result column as 'Total Salary'.
- The GROUP BY clause groups the result set by the 'department_id' column, ensuring that calculations are performed for each unique department.
- The query does not filter or manipulate the data further; it simply retrieves the department code, the number of employees, and the total salary for each department from the 'employees' table.
Output:
Department Code No of Employees Total Salary
--------------- --------------- ------------
100 6 51608
30 6 24900
1 7000
90 3 58000
20 2 19000
70 1 10000
110 2 20308
50 45 156400
80 34 304500
40 1 6500
60 5 28800
10 1 4400
SQL GROUP BY on more than one columns
The following query displays the department code, job id, total salary paid to employees group by department_id, job_id.
Sample table: employees
employee_id first_name last_name email phone_number hire_date job_id salary commission_pct manager_id department_id ----------- ---------- ---------- ---------- ------------ ---------- ---------- ---------- -------------- ---------- ------------- 100 Steven King SKING 515.123.4567 6/17/1987 AD_PRES 24000 90 101 Neena Kochhar NKOCHHAR 515.123.4568 6/18/1987 AD_VP 17000 100 90 102 Lex De Haan LDEHAAN 515.123.4569 6/19/1987 AD_VP 17000 100 90 103 Alexander Hunold AHUNOLD 590.423.4567 6/20/1987 IT_PROG 9000 102 60 104 Bruce Ernst BERNST 590.423.4568 6/21/1987 IT_PROG 6000 103 60 105 David Austin DAUSTIN 590.423.4569 6/22/1987 IT_PROG 4800 103 60 106 Valli Pataballa VPATABAL 590.423.4560 6/23/1987 IT_PROG 4800 103 60 107 Diana Lorentz DLORENTZ 590.423.5567 6/24/1987 IT_PROG 4200 103 60 108 Nancy Greenberg NGREENBE 515.124.4569 6/25/1987 FI_MGR 12000 101 100 109 Daniel Faviet DFAVIET 515.124.4169 6/26/1987 FI_ACCOUNT 9000 108 100 ........... 206 William Gietz WGIETZ 515.123.8181 10/1/1987 AC_ACCOUNT 8300 205 110
SQL Code:
-- This SQL query calculates the total salary for each combination of department and job, and displays the department code, job ID, and total salary.
-- SELECT statement begins
SELECT
department_id "Department Code", -- Selects the 'department_id' column and renames it as 'Department Code' in the result set
job_id, -- Selects the 'job_id' column
SUM(salary) "Total Salary" -- Calculates the sum of salaries for each combination of department and job, and renames the result column as 'Total Salary'
FROM
employees -- Specifies the 'employees' table to retrieve data from
GROUP BY
department_id, -- Groups the result set by the 'department_id' column
job_id; -- Groups the result set further by the 'job_id' column
Explanation:
- This SQL code is a SELECT statement that retrieves data from the 'employees' table.
- It calculates the total salary for each combination of department and job and displays the department code, job ID, and total salary.
- The SELECT clause selects the 'department_id' column from the 'employees' table and renames it as 'Department Code' in the result set.
- It also selects the 'job_id' column to display the job ID in the result set.
- The SUM(salary) function calculates the sum of salaries for each combination of department and job.
- The GROUP BY clause groups the result set first by the 'department_id' column and then further by the 'job_id' column, ensuring that calculations are performed for each unique combination.
- The query retrieves the department code, job ID, and total salary for each combination of department and job from the 'employees' table.
Output:
Department Code JOB_ID Total Salary
--------------- ---------- ------------
110 AC_ACCOUNT 8300
90 AD_VP 34000
50 ST_CLERK 55700
80 SA_REP 243500
50 ST_MAN 36400
80 SA_MAN 61000
110 AC_MGR 12008
90 AD_PRES 24000
60 IT_PROG 28800
100 FI_MGR 12008
30 PU_CLERK 13900
50 SH_CLERK 64300
20 MK_MAN 13000
100 FI_ACCOUNT 39600
SA_REP 7000
70 PR_REP 10000
30 PU_MAN 11000
10 AD_ASST 4400
20 MK_REP 6000
40 HR_REP 6500
SQL GROUP BY with WHERE clause
The following query displays the department code, total salary paid to employees group by department_id and manager_id=103.
Sample table: employees
employee_id first_name last_name email phone_number hire_date job_id salary commission_pct manager_id department_id ----------- ---------- ---------- ---------- ------------ ---------- ---------- ---------- -------------- ---------- ------------- 100 Steven King SKING 515.123.4567 6/17/1987 AD_PRES 24000 90 101 Neena Kochhar NKOCHHAR 515.123.4568 6/18/1987 AD_VP 17000 100 90 102 Lex De Haan LDEHAAN 515.123.4569 6/19/1987 AD_VP 17000 100 90 103 Alexander Hunold AHUNOLD 590.423.4567 6/20/1987 IT_PROG 9000 102 60 104 Bruce Ernst BERNST 590.423.4568 6/21/1987 IT_PROG 6000 103 60 105 David Austin DAUSTIN 590.423.4569 6/22/1987 IT_PROG 4800 103 60 106 Valli Pataballa VPATABAL 590.423.4560 6/23/1987 IT_PROG 4800 103 60 107 Diana Lorentz DLORENTZ 590.423.5567 6/24/1987 IT_PROG 4200 103 60 108 Nancy Greenberg NGREENBE 515.124.4569 6/25/1987 FI_MGR 12000 101 100 109 Daniel Faviet DFAVIET 515.124.4169 6/26/1987 FI_ACCOUNT 9000 108 100 ........... 206 William Gietz WGIETZ 515.123.8181 10/1/1987 AC_ACCOUNT 8300 205 110
SQL Code:
-- This SQL query calculates the total salary for each department where the manager ID is 103, and displays the department code and total salary.
-- SELECT statement begins
SELECT
department_id "Department Code", -- Selects the 'department_id' column and renames it as 'Department Code' in the result set
SUM(salary) "Total Salary" -- Calculates the sum of salaries for each department where the manager ID is 103, and renames the result column as 'Total Salary'
FROM
employees -- Specifies the 'employees' table to retrieve data from
WHERE
MANAGER_ID = 103 -- Filters the rows where the manager ID is 103
GROUP BY
department_id; -- Groups the result set by the 'department_id' column
Explanation:
- This SQL code is a SELECT statement that retrieves data from the 'employees' table.
- It calculates the total salary for each department where the manager ID is 103 and displays the department code and total salary.
- The SELECT clause selects the 'department_id' column from the 'employees' table and renames it as 'Department Code' in the result set.
- The SUM(salary) function calculates the sum of salaries for each department where the manager ID is 103.
- The WHERE clause filters the rows where the manager ID is 103.
- The GROUP BY clause groups the result set by the 'department_id' column, ensuring that calculations are performed for each unique department.
- The query retrieves the department code and total salary for each department where the manager ID is 103 from the 'employees' table.
Output:
Department Code Total Salary
--------------- ------------
60 19800
SQL GROUP BY with HAVING clause
The following query displays the department id, number of employees of those groups that have more than 2 employees:
Sample table: employees
employee_id first_name last_name email phone_number hire_date job_id salary commission_pct manager_id department_id ----------- ---------- ---------- ---------- ------------ ---------- ---------- ---------- -------------- ---------- ------------- 100 Steven King SKING 515.123.4567 6/17/1987 AD_PRES 24000 90 101 Neena Kochhar NKOCHHAR 515.123.4568 6/18/1987 AD_VP 17000 100 90 102 Lex De Haan LDEHAAN 515.123.4569 6/19/1987 AD_VP 17000 100 90 103 Alexander Hunold AHUNOLD 590.423.4567 6/20/1987 IT_PROG 9000 102 60 104 Bruce Ernst BERNST 590.423.4568 6/21/1987 IT_PROG 6000 103 60 105 David Austin DAUSTIN 590.423.4569 6/22/1987 IT_PROG 4800 103 60 106 Valli Pataballa VPATABAL 590.423.4560 6/23/1987 IT_PROG 4800 103 60 107 Diana Lorentz DLORENTZ 590.423.5567 6/24/1987 IT_PROG 4200 103 60 108 Nancy Greenberg NGREENBE 515.124.4569 6/25/1987 FI_MGR 12000 101 100 109 Daniel Faviet DFAVIET 515.124.4169 6/26/1987 FI_ACCOUNT 9000 108 100 ........... 206 William Gietz WGIETZ 515.123.8181 10/1/1987 AC_ACCOUNT 8300 205 110
SQL Code:
-- This SQL query counts the number of employees in each department and displays the department ID along with the count,
-- but only for departments with more than two employees.
-- SELECT statement begins
SELECT
department_id, -- Selects the 'department_id' column
count(*) "No. of Employee" -- Counts the number of rows for each department and renames the result column as 'No. of Employee'
FROM
employees -- Specifies the 'employees' table to retrieve data from
GROUP BY
department_id -- Groups the result set by the 'department_id' column
HAVING
count(*) > 2; -- Filters the grouped results to include only departments with more than two employees
Explanation:
- This SQL code is a SELECT statement that retrieves data from the 'employees' table.
- It counts the number of employees in each department and displays the department ID along with the count.
- The SELECT clause selects the 'department_id' column from the 'employees' table.
- The count(*) function counts the number of rows for each department.
- The GROUP BY clause groups the result set by the 'department_id' column, ensuring that calculations are performed for each unique department.
- The HAVING clause filters the grouped results to include only departments with more than two employees.
- The query retrieves the department ID and the count of employees for each department from the 'employees' table, but only for departments with more than two employees.
Output:
DEPARTMENT_ID No. of Employee
------------- ---------------
100 6
30 6
90 3
50 45
80 34
60 5
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