SQL Exercise: Employees who work in any department located in London
24. From the following tables, write a SQL query to find full name (first and last name), and salary of all employees working in any department in the city of London.
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 | | 100 | Finance | 108 | 1700 | ...... | 270 | Payroll | 0 | 1700 | +---------------+----------------------+------------+-------------+
Sample table: locations
+-------------+------------------------------------------+-------------+---------------------+-------------------+------------+ | LOCATION_ID | STREET_ADDRESS | POSTAL_CODE | CITY | STATE_PROVINCE | COUNTRY_ID | +-------------+------------------------------------------+-------------+---------------------+-------------------+------------+ | 1000 | 1297 Via Cola di Rie | 989 | Roma | | IT | | 1100 | 93091 Calle della Testa | 10934 | Venice | | IT | | 1200 | 2017 Shinjuku-ku | 1689 | Tokyo | Tokyo Prefecture | JP | | 1300 | 9450 Kamiya-cho | 6823 | Hiroshima | | JP | | 1400 | 2014 Jabberwocky Rd | 26192 | Southlake | Texas | US | | 1500 | 2011 Interiors Blvd | 99236 | South San Francisco | California | US | | 1600 | 2007 Zagora St | 50090 | South Brunswick | New Jersey | US | | 1700 | 2004 Charade Rd | 98199 | Seattle | Washington | US | | 1800 | 147 Spadina Ave | M5V 2L7 | Toronto | Ontario | CA | | 1900 | 6092 Boxwood St | YSW 9T2 | Whitehorse | Yukon | CA | ........ | 3200 | Mariano Escobedo 9991 | 11932 | Mexico City | Distrito Federal, | MX | +-------------+------------------------------------------+-------------+---------------------+-------------------+------------+
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 | 2003-06-17 | AD_PRES | 24000.00 | 0.00 | 0 | 90 | | 101 | Neena | Kochhar | NKOCHHAR | 515.123.4568 | 2005-09-21 | AD_VP | 17000.00 | 0.00 | 100 | 90 | | 102 | Lex | De Haan | LDEHAAN | 515.123.4569 | 2001-01-13 | AD_VP | 17000.00 | 0.00 | 100 | 90 | | 103 | Alexander | Hunold | AHUNOLD | 590.423.4567 | 2006-01-03 | IT_PROG | 9000.00 | 0.00 | 102 | 60 | | 104 | Bruce | Ernst | BERNST | 590.423.4568 | 2007-05-21 | IT_PROG | 6000.00 | 0.00 | 103 | 60 | | 105 | David | Austin | DAUSTIN | 590.423.4569 | 2005-06-25 | IT_PROG | 4800.00 | 0.00 | 103 | 60 | | 106 | Valli | Pataballa | VPATABAL | 590.423.4560 | 2006-02-05 | IT_PROG | 4800.00 | 0.00 | 103 | 60 | | 107 | Diana | Lorentz | DLORENTZ | 590.423.5567 | 2007-02-07 | IT_PROG | 4200.00 | 0.00 | 103 | 60 | | 108 | Nancy | Greenberg | NGREENBE | 515.124.4569 | 2002-08-17 | FI_MGR | 12008.00 | 0.00 | 101 | 100 | | 109 | Daniel | Faviet | DFAVIET | 515.124.4169 | 2002-08-16 | FI_ACCOUNT | 9000.00 | 0.00 | 108 | 100 | ...... | 206 | William | Gietz | WGIETZ | 515.123.8181 | 2002-06-07 | AC_ACCOUNT | 8300.00 | 0.00 | 205 | 110 | +-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+
Sample Solution:
-- Selecting specific columns (first_name || ' ' || last_name AS Employee_name, salary) from the 'employees' table
SELECT first_name || ' ' || last_name AS Employee_name, salary
-- Performing an INNER JOIN between the 'employees' table and the 'departments' table using the common column 'department_id'
FROM employees
JOIN departments USING (department_id)
-- Performing another INNER JOIN between the result set and the 'locations' table using the common column 'location_id'
JOIN locations USING (location_id)
-- Filtering rows based on the condition that 'city' is equal to 'London'
WHERE city = 'London';
Sample Output:
employee_name salary Susan Mavris 6500.00
Code Explanation:
The said query in SQL which will return a list of employee names, along with their salaries, for employees who work in a department located in the city of London, based on data from the employees, departments, and locations tables.
JOIN clause joins employees with departments and then joins locations with location_id.
The WHERE clause filters the results to only include employees who work in a department located in the city of London.
Alternative Solution:
Using INNER JOIN with Explicit Column Names and WHERE Clause:
SELECT employees.first_name || ' ' || employees.last_name AS Employee_name, employees.salary
FROM employees
JOIN departments ON employees.department_id = departments.department_id
JOIN locations ON departments.location_id = locations.location_id
WHERE locations.city = 'London';
Explanation:
This query performs INNER JOINs between 'employees', 'departments', and 'locations' tables based on matching keys (department_id and location_id). It retrieves the concatenated full name of employees and their salaries, filtered by the condition that the city is 'London'.
Go to:
PREV : Number of days worked for all jobs in department 80.
NEXT : Employees worked without a commission percentage.
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Query Visualization:
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