EasySQL challenge #76GROUP BY

Reviews per Product

Count number of reviews for each product.

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

Solve This SQL Challenge

The task

Count number of reviews for each product.

A correct answer returns 93 rows with 2 columns: product_id, COUNT(*).

Tables used in this challenge

`reviews` 180 rows

id INTEGER (PK) · product_id INTEGER · user_id INTEGER · rating INTEGER · comment TEXT · reviewed_at TEXT · helpful_votes INTEGER

First 3 rows of the reviews table
id	product_id	user_id	rating	comment	reviewed_at	helpful_votes
1	23	45	5	Works perfectly. Very happy with this purchase.	2022-09-02	40
2	101	68	5	Excellent build quality. Will buy again!	2023-01-05	96
3	40	39	1	Would give 6 stars if I could!	2023-05-15	17

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	COUNT(*)
1	2
2	3
3	1
4	1
5	1

First 5 of 93 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 product_id

Show the step-by-step approach

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

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

Model solution

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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(*).