HardSQL challenge #375SELF JOINDATEAGGREGATION

Max Orders in 14-Day Window

Find users with highest order count in any 14-day period.

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

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The task

Find users with highest order count in any 14-day period.

A correct answer returns 200 rows with 2 columns: user_id, cnt.

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

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
user_idcnt
472
1242
11
21
41

First 5 of 200 rows. This is the output of the model solution on the practice database.

What you practise

  • Filtering a range of values with BETWEEN
  • 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
  • Joining a table to itself under two aliases
  • Working with dates in SQLite: strftime, julianday and DATE
  • Sorting the result with ORDER BY

Hint and approach

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

Show the hint

self join date range

Show the step-by-step approach

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

  1. Read from the orders table, under the alias o1.
  2. Join orders again, under the alias o2, on o1.user_id = o2.user_id, and only where o2.order_date is between o1.order_date and 14 days after o1.order_date, both ends included.
  3. Group the rows by o1.user_id and o1.order_date, so the result has one row for each combination.
  4. For each group, compute COUNT(*).
  5. Sort the result by cnt descending.

Model solution

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What to watch for

  • Self join. Without a condition such as a.id < b.id, a self join pairs every row with itself and lists each pair twice.
  • 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.
  • GROUP BY on several columns. Each extra GROUP BY column splits the groups further. Grouping by a column that is unique per row gives back one group per row.
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