HardSQL challenge #240WINDOW FUNCTIONPARTITIONAGGREGATION

Find the Top 2 Products by Revenue in Each Category

Return the two highest-revenue products of every category.

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

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The task

Find top 2 products by total revenue in each category.

A correct answer returns 20 rows with 4 columns: category, id, revenue, rn.

Tables used in this challenge

products 120 rows

id INTEGER (PK) · name TEXT · category TEXT · price REAL · cost REAL · stock INTEGER · sku TEXT · status TEXT · rating REAL · review_count INTEGER

First 3 rows of the products table
idnamecategorypricecoststockskustatusratingreview_count
1Play Kitchen DeluxeToys1786.38589.51257SKU-000001active457
2Micro-Cut Paper Shredder 8-SheetOffice Supplies907.93281.46131SKU-000002active2.5447
34K Dual Dash CameraAutomotive568.89204.8176SKU-000003active4.129

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
categoryidrevenuern
Automotive607519.961
Automotive726460.592
Books395027.81
Books1054214.452
Clothing1198995.831

First 5 of 20 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
  • Summarising many rows into one value with COUNT, SUM, AVG, MIN or MAX
  • Grouping by more than one column to get a row per combination
  • Combining rows from two tables on a matching key with an inner join
  • Querying the result of another query with a subquery in FROM
  • Numbering rows with ROW_NUMBER() to pick one or a few per group
  • Restarting a window calculation for each group with PARTITION BY

Hint and approach

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

Show the hint

SUM + ROW_NUMBER PARTITION

Show the step-by-step approach

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

  1. Build an inner query first (a derived table):
    1. Read from the products table.
    2. Join order_items on products.id = order_items.product_id.
    3. Group the rows by products.category and products.id, so the result has one row for each combination.
    4. For each group, compute SUM(order_items.subtotal).
    5. Number the rows with ROW_NUMBER(), ordered by SUM(order_items.subtotal) descending, restarting for each products.category.
  2. Filter in WHERE: keep the rows where rn is at most 2. The filter sits in the outer query because a window function cannot be tested in the WHERE clause of the query that computes it.

Model solution

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What to watch for

Revenue is summed per product by GROUP BY, the window function numbers those sums inside each category, and the outer query keeps numbers 1 and 2. Grouping by both category and product id is what leaves one row per product.

  • ROW_NUMBER(). ROW_NUMBER() breaks ties arbitrarily. When the ORDER BY inside OVER does not identify one row, the row that gets number 1 can change between runs.
  • PARTITION BY. PARTITION BY is not GROUP BY: it keeps every row and only decides which rows a window function looks at.
  • Derived table. Columns computed in the inner query are only visible outside under their alias, so every calculated column there needs a name.
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Same technique, other tables

More practice on this pattern with different columns:

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