MediumSQL challenge #170JOINGROUP BY

Count Orders and Payments per User Without Double Counting

Return every user with their number of orders and their number of payments.

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

Solve This SQL Challenge

The task

Find total orders and payments per user.

A correct answer returns 150 rows with 3 columns: id, orders, payments.

Tables used in this challenge

`users` 150 rows

id INTEGER (PK) · first_name TEXT · last_name TEXT · email TEXT · phone TEXT · city TEXT · country TEXT · created_at TEXT · status TEXT · is_premium INTEGER

First 3 rows of the users table
id	first_name	last_name	email	phone	city	country	created_at	status
1	Jennifer	Torres	jennifer.torres57@icloud.com	+1-263-376-2371	Houston	Canada	2024-04-13	active
2	Donna	Adams	donna.adams52@gmail.com	+1-753-780-6670	Fort Worth	Canada	2020-09-15	inactive
3	Kenneth	Young	kenneth.young94@proton.me	+1-440-743-9503	Austin	USA	2023-09-05	active

`orders` 200 rows

id INTEGER (PK) · user_id INTEGER · total REAL · discount REAL · item_count INTEGER · status TEXT · order_date TEXT · shipped_date TEXT · shipping_city TEXT · shipping_country TEXT

First 3 rows of the orders table
id	user_id	total	discount	item_count	status	order_date	shipped_date	shipping_city	shipping_country
1	43	2638.11	0.25	3	cancelled	2024-04-24	NULL	Los Angeles	USA
2	8	2947	0.19	5	delivered	2024-03-15	2023-04-09	Chicago	USA
3	136	683.24	0.08	2	delivered	2023-02-15	2023-07-10	Los Angeles	USA

`payments` 200 rows

id INTEGER (PK) · order_id INTEGER · user_id INTEGER · amount REAL · method TEXT · status TEXT · paid_at TEXT · transaction_id TEXT

First 3 rows of the payments table
id	order_id	user_id	amount	method	status	paid_at	transaction_id
1	46	105	921.62	Gift Card	success	2022-05-18	TXNHTN4GIK5L
2	157	37	1723.82	Bank Transfer	success	2024-11-11	TXNKJPIXJOFP
3	45	36	978.76	Bank Transfer	refunded	2023-10-24	TXNU2RW4MK1R

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
id	orders	payments
1	1	0
2	1	1
3	0	1
4	3	0
5	1	0

First 5 of 150 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
Counting different values with COUNT(DISTINCT …)
Grouping rows with GROUP BY to get one result row per group
Keeping every row of the left table with LEFT JOIN, matched or not

Hint and approach

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

Show the hint

Two LEFT JOINs multiply rows, so count each side with COUNT(DISTINCT ...)

Show the step-by-step approach

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

Read from the users table.
LEFT JOIN orders on users.id = orders.user_id. Rows from the left side that have no match are kept, with NULL in the joined columns.
LEFT JOIN payments on users.id = payments.user_id. Rows from the left side that have no match are kept, with NULL in the joined columns.
Group the rows by users.id, so the result has one row for each value.
For each group, compute COUNT(DISTINCT orders.id), which counts each different value once and COUNT(DISTINCT payments.id), which counts each different value once.

Model solution

Checking your session…

What to watch for

Joining users to orders and to payments in one query pairs every order of a user with every payment of that user. A user with 3 orders and 2 payments becomes 6 rows, so COUNT(o.id) would say 6. COUNT(DISTINCT …) on each side counts each order and each payment once.

LEFT JOIN. A WHERE condition on the right-hand table removes the unmatched rows and turns the LEFT JOIN back into an inner join. Conditions on that table go in the ON clause.
COUNT(DISTINCT). After a join, COUNT(*) counts joined rows. Counting the parent entity needs COUNT(DISTINCT its id).
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.