MediumSQL challenge #166GROUP BY

Count Distinct Products per Order

Return each order with the number of different products it contains.

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

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The task

Find distinct products count per order.

A correct answer returns 200 rows with 2 columns: order_id, COUNT(DISTINCT product_id).

Tables used in this challenge

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
idorder_idproduct_idquantityunit_pricesubtotal
11765192.21961.05
21823107.1321.3
316515.395.39

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
order_idCOUNT(DISTINCT product_id)
13
21
32
41
54

First 5 of 200 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

Hint and approach

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

Show the hint

COUNT DISTINCT

Show the step-by-step approach

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

  1. Read from the order_items table.
  2. Group the rows by order_id, so the result has one row for each value.
  3. For each group, compute COUNT(DISTINCT product_id), which counts each different value once.

Model solution

Checking your session…

What to watch for

COUNT(DISTINCT product_id) counts each product once per order even if it appears on two lines. COUNT(*) would count the lines instead.

  • 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.
  • Aggregate functions. Aggregates skip NULLs. AVG divides by the number of non-NULL values and COUNT(column) can be smaller than COUNT(*).
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Same technique, other tables

More practice on this pattern with different columns:

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