MediumSQL challenge #159GROUP BY

Employees per City

Count employees grouped by city.

The challenge runs on the SQLabHub practice database (SQLite) and uses the employees table. You can solve it in the browser, with nothing to install.

Solve This SQL Challenge

The task

Count employees grouped by city.

A correct answer returns 20 rows with 2 columns: city, COUNT(*).

Tables used in this challenge

`employees` 120 rows

id INTEGER (PK) · first_name TEXT · last_name TEXT · email TEXT · department TEXT · department_id INTEGER · title TEXT · salary REAL · hire_date TEXT · manager_id INTEGER · city TEXT · status TEXT

First 3 rows of the employees table
id	first_name	last_name	email	department	department_id	title	salary	hire_date	manager_id	city	status
1	William	Lewis	wlewis18@corp.io	Customer Success	7	Senior CS Manager	61774	2022-07-24	NULL	San Diego	active
2	John	Miller	jmiller51@corp.io	Legal	8	Paralegal	191694	2021-08-09	NULL	San Diego	active
3	Daniel	Rivera	drivera50@corp.io	Engineering	1	VP Engineering	135067	2016-06-06	NULL	Houston	active

The first rows of each table are shown. See how all the tables relate on the database schema page.

Expected output

Expected output of the challenge
city	COUNT(*)
Austin	4
Charlotte	4
Chicago	6
Columbus	8
Dallas	2

First 5 of 20 rows. This is the output of the model solution on the practice database. The task does not fix a sort order, so your rows may come back in a different order.

What you practise

Summarising many rows into one value with COUNT, SUM, AVG, MIN or MAX
Grouping rows with GROUP BY to get one result row per group

Hint and approach

Try the challenge first. Open these only when you need them.

Show the hint

GROUP BY city

Show the step-by-step approach

How the model solution is built, in the order the database evaluates it:

Read from the employees table.
Group the rows by city, so the result has one row for each value.
For each group, compute COUNT(*).

Model solution

Checking your session…

What to watch for

GROUP BY. Every selected column must either be in GROUP BY or sit inside an aggregate. SQLite accepts other columns and returns an arbitrary value for them; most databases raise an error.
Aggregate functions. Aggregates skip NULLs. AVG divides by the number of non-NULL values and COUNT(column) can be smaller than COUNT(*).