SQL Exercise: Personnel with department ID 2001 and ID 1001
SQL subqueries on employee Database: Exercise-10 with Solution
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10. From the following table, write a SQL query to find those employees of department ID 2001 and whose designation is same as of the designation of department ID 1001. Return complete information about the employees.
Sample table: employees
Sample table: department
Sample Solution:
SELECT *
FROM employees e,
department d
WHERE d.dep_id = 2001
AND e.dep_id = d.dep_id
AND e.job_name IN
(SELECT e.job_name
FROM employees e,
department d
WHERE e.dep_id = d.dep_id
AND d.dep_id =1001);
Sample Output:
emp_id | emp_name | job_name | manager_id | hire_date | salary | commission | dep_id | dep_id | dep_name | dep_location --------+----------+----------+------------+------------+---------+------------+--------+--------+----------+-------------- 65646 | JONAS | MANAGER | 68319 | 1991-04-02 | 2957.00 | | 2001 | 2001 | AUDIT | MELBOURNE 63679 | SANDRINE | CLERK | 69062 | 1990-12-18 | 900.00 | | 2001 | 2001 | AUDIT | MELBOURNE 68736 | ADNRES | CLERK | 67858 | 1997-05-23 | 1200.00 | | 2001 | 2001 | AUDIT | MELBOURNE (3 rows)
Explanation:
The given query in SQL that selects all employees who belong to department ID 2001 and whose job_name is also found in the job_names of employees who belong to department ID 1001 from the 'employees' and 'department' tables.
The WHERE clause in the main query filters the results to only include rows where the department ID is 2001 and the employee's department ID is also 2001 and the job_name of the employee must be found in the job_names of employees who belong to department ID 1001 which is determined by the subquery.
The subquery selects the job_name of all employees who belong to department ID 1001 by joining the 'employees' and 'department' tables.
The subquery is used in the main query as a condition for the IN operator.
Practice Online
Structure of employee Database:
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Previous SQL Exercise: List the employees who are senior to ADELYN.
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