SQL Exercise: List the number of employees and average salary
SQL employee Database: Exercise-104 with Solution
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104. From the following table, write a SQL query to find number of employees and average salary. Group the result set on department id and job name. Return number of employees, average salary, department ID, and job name.
Sample table: employees
Sample Solution:
SELECT count(*),
avg(salary),
dep_id,
job_name
FROM employees
GROUP BY dep_id,
job_name;
Sample Output:
count | avg | dep_id | job_name -------+-----------------------+--------+----------- 1 | 2750.0000000000000000 | 3001 | MANAGER 2 | 3100.0000000000000000 | 2001 | ANALYST 4 | 1500.0000000000000000 | 3001 | SALESMAN 1 | 2550.0000000000000000 | 1001 | MANAGER 1 | 6000.0000000000000000 | 1001 | PRESIDENT 1 | 2957.0000000000000000 | 2001 | MANAGER 2 | 1050.0000000000000000 | 2001 | CLERK 1 | 1400.0000000000000000 | 1001 | CLERK 1 | 1050.0000000000000000 | 3001 | CLERK (9 rows)
Explanation:
The said query in SQL that returns the number of employees, average salary, department ID, and job name for each unique combination of dep_id and job_name from the employees table.
The GROUP BY CLAUSE groups the results by dep_id and job_name columns.
Relational Algebra Expression:

Relational Algebra Tree:

Practice Online
Sample Database: employee

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Previous SQL Exercise: Check if all the employees numbers are indeed unique.
Next SQL Exercise: List the names of those employees starting with A.
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SQL: Tips of the Day
Grouped LIMIT in PostgreSQL: Show the first N rows for each group?
db=# SELECT * FROM xxx; id | section_id | name ----+------------+------ 1 | 1 | A 2 | 1 | B 3 | 1 | C 4 | 1 | D 5 | 2 | E 6 | 2 | F 7 | 3 | G 8 | 2 | H (8 rows)
I need the first 2 rows (ordered by name) for each section_id, i.e. a result similar to:
id | section_id | name ----+------------+------ 1 | 1 | A 2 | 1 | B 5 | 2 | E 6 | 2 | F 7 | 3 | G (5 rows)
PostgreSQL v9.3 you can do a lateral join
select distinct t_outer.section_id, t_top.id, t_top.name from t t_outer join lateral ( select * from t t_inner where t_inner.section_id = t_outer.section_id order by t_inner.name limit 2 ) t_top on true order by t_outer.section_id;
Database: PostgreSQL
Ref: https://bit.ly/3AfYwZI
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