PostgreSQL JOINS: Make a join with tables employees and departments to get the department name and number of employees working in each department
6. Write a query to make a join with two tables employees and departments to get the department name and number of employees working in each department.
Sample Solution:
Code:
-- This SQL query retrieves the department name along with the count of employees in each department, ordered alphabetically by department name.
SELECT department_name AS "Department Name", -- Selects the department_name column and labels it as "Department Name"
COUNT(*) AS "No of Employees" -- Calculates the count of employees in each department and labels it as "No of Employees"
FROM departments -- Specifies the first table from which to retrieve data, in this case, the departments table
INNER JOIN employees -- Performs an inner join between the departments and employees tables
ON employees.department_id = departments.department_id -- Specifies the join condition based on the department_id column
GROUP BY departments.department_id, department_name -- Groups the results by department_id and department_name
ORDER BY department_name; -- Orders the results alphabetically by department_name
Explanation:
- This SQL query retrieves the department name along with the count of employees in each department, ordered alphabetically by department name.
- The SELECT statement selects the department_name column from the departments table and calculates the count of employees in each department, labeling them as "Department Name" and "No of Employees" respectively.
- The FROM clause specifies the first table from which to retrieve data, which is the departments table.
- An INNER JOIN operation is performed between the departments and employees tables based on the common column department_id.
- The ON clause specifies the join condition where the department_id in the employees table matches the department_id in the departments table.
- The GROUP BY clause groups the results by department_id and department_name.
- The ORDER BY clause orders the results alphabetically by department_name.
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 | 1987-06-17 | AD_PRES | 24000.00 | 0.00 | 0 | 90 | | 101 | Neena | Kochhar | NKOCHHAR | 515.123.4568 | 1987-06-18 | AD_VP | 17000.00 | 0.00 | 100 | 90 | | 102 | Lex | De Haan | LDEHAAN | 515.123.4569 | 1987-06-19 | AD_VP | 17000.00 | 0.00 | 100 | 90 | | 103 | Alexander | Hunold | AHUNOLD | 590.423.4567 | 1987-06-20 | IT_PROG | 9000.00 | 0.00 | 102 | 60 | | 104 | Bruce | Ernst | BERNST | 590.423.4568 | 1987-06-21 | IT_PROG | 6000.00 | 0.00 | 103 | 60 | | 105 | David | Austin | DAUSTIN | 590.423.4569 | 1987-06-22 | IT_PROG | 4800.00 | 0.00 | 103 | 60 | | 106 | Valli | Pataballa | VPATABAL | 590.423.4560 | 1987-06-23 | IT_PROG | 4800.00 | 0.00 | 103 | 60 | | 107 | Diana | Lorentz | DLORENTZ | 590.423.5567 | 1987-06-24 | IT_PROG | 4200.00 | 0.00 | 103 | 60 | | 108 | Nancy | Greenberg | NGREENBE | 515.124.4569 | 1987-06-25 | FI_MGR | 12000.00 | 0.00 | 101 | 100 | .......... | 206 | William | Gietz | WGIETZ | 515.123.8181 | 1987-10-01 | AC_ACCOUNT | 8300.00 | 0.00 | 205 | 110 | +-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+
Sample table: departments
+---------------+----------------------+------------+-------------+ | DEPARTMENT_ID | DEPARTMENT_NAME | MANAGER_ID | LOCATION_ID | +---------------+----------------------+------------+-------------+ | 10 | Administration | 200 | 1700 | | 20 | Marketing | 201 | 1800 | | 30 | Purchasing | 114 | 1700 | | 40 | Human Resources | 203 | 2400 | | 50 | Shipping | 121 | 1500 | | 60 | IT | 103 | 1400 | | 70 | Public Relations | 204 | 2700 | | 80 | Sales | 145 | 2500 | | 90 | Executive | 100 | 1700 | ........ | 270 | Payroll | 0 | 1700 | +---------------+----------------------+------------+-------------+
Output:
pg_exercises=# SELECT department_name AS "Department Name", pg_exercises-# COUNT(*) AS "No of Employees" pg_exercises-# FROM departments pg_exercises-# INNER JOIN employees pg_exercises-# ON employees.department_id = departments.department_id pg_exercises-# GROUP BY departments.department_id, department_name pg_exercises-# ORDER BY department_name; Department Name | No of Employees ------------------+----------------- Accounting | 2 Administration | 1 Executive | 3 Finance | 6 Human Resources | 1 IT | 5 Marketing | 2 Public Relations | 1 Purchasing | 6 Sales | 33 Shipping | 45 (11 rows)
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PREV : Write a query to make a join with a table employees and itself to find the name, including first_name and last_name and hire date for those employees who were hired after the employee Jones.
NEXT : Write a query to make a join to find the employee ID, job title and number of days an employee worked, for all the employees who worked in a department which ID is 90.
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