MediumSQL challenge #177GROUP BYHAVING

Employees Sharing Same Manager

Find employees who share the same manager.

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

Find employees who share the same manager.

A correct answer returns 10 rows with 2 columns: manager_id, 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
manager_id	COUNT(*)
1	11
2	13
3	11
4	11
5	10
6	2
7	13
8	12
9	14
10	13

The complete result: 10 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

Filtering rows with a WHERE condition
Testing for missing values with IS NULL and IS NOT NULL
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
Filtering groups after aggregation with HAVING

Hint and approach

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

Show the hint

GROUP BY manager_id

Show the step-by-step approach

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

Read from the employees table.
Filter in WHERE: keep the rows where manager_id has a value (it is not NULL).
Group the rows by manager_id, so the result has one row for each value.
For each group, compute COUNT(*).
Filter the groups in HAVING: keep those where COUNT(*) is greater than 1.

Model solution

Checking your session…

What to watch for

HAVING. Conditions on single rows belong in WHERE and conditions on aggregates in HAVING. Putting a row filter in HAVING works, but it groups rows that are about to be thrown away.
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.
IS NULL. Comparing with = NULL or <> NULL never matches any row. NULL has to be tested with IS NULL or IS NOT NULL.