HardSQL challenge #378JOINGROUP BYANALYSIS

Highest Retention Products

Find products with most repeat buyers over time.

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

Solve This SQL Challenge

The task

Find products with most repeat buyers over time.

A correct answer returns 6 rows with 2 columns: product_id, repeat_users.

Tables used in this challenge

`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

`order_items` 496 rows

id INTEGER (PK) · order_id INTEGER · product_id INTEGER · quantity INTEGER · unit_price REAL · subtotal REAL

First 3 rows of the order_items table
id	order_id	product_id	quantity	unit_price	subtotal
1	1	76	5	192.21	961.05
2	1	82	3	107.1	321.3
3	1	65	1	5.39	5.39

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
product_id	repeat_users
113	1
65	1
59	1
29	1
23	1
5	1

The complete result: 6 rows. This is the output of the model solution on the practice database.

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
Grouping by more than one column to get a row per combination
Filtering groups after aggregation with HAVING
Combining rows from two tables on a matching key with an inner join
Querying the result of another query with a subquery in FROM
Sorting the result with ORDER BY

Hint and approach

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

Show the hint

COUNT DISTINCT user_id with repeats

Show the step-by-step approach

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

Build an inner query first (a derived table):
1. Read from the order_items table.
2. Join orders on order_items.order_id = orders.id.
3. Group the rows by order_items.product_id and orders.user_id, so the result has one row for each combination.
4. For each group, compute COUNT(*).
5. Filter the groups in HAVING: keep those where COUNT(*) is greater than 1.
Group the rows by product_id, so the result has one row for each value.
For each group, compute COUNT(*).
Sort the result by repeat_users descending.

Model solution

Checking your session…

What to watch for

Derived table. Columns computed in the inner query are only visible outside under their alias, so every calculated column there needs a name.
INNER JOIN. An inner join drops rows that have no partner on the other side, and it repeats a row once for every match, which inflates SUM and COUNT.
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.