HardSQL challenge #320WINDOW FUNCTIONGROUP BYPARTITION

Most Loyal Buyer per Product

Find user who bought each product most frequently.

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.

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The task

Find user who bought each product most frequently.

A correct answer returns 119 rows with 4 columns: product_id, user_id, cnt, rn.

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
iduser_idtotaldiscountitem_countstatusorder_dateshipped_dateshipping_cityshipping_country
1432638.110.253cancelled2024-04-24NULLLos AngelesUSA
2829470.195delivered2024-03-152023-04-09ChicagoUSA
3136683.240.082delivered2023-02-152023-07-10Los AngelesUSA

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
product_iduser_idcntrn
13711
211411
311211
49111
57721

First 5 of 119 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

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 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. Number the rows with ROW_NUMBER(), ordered by COUNT(*) descending, restarting for each order_items.product_id.
  2. Filter in WHERE: keep the rows where rn equals 1. 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

Checking your session…

What to watch for

  • 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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