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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 Expression: List the no. of employees and average salary within each department for each job name.

Relational Algebra Tree:

Relational Algebra Tree: List the no. of employees and average salary within each department for each job name.

Practice Online


Sample Database: employee

employee database structure

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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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