HardSQL challenge #265DATEGROUP BYORDER BY

Top Users in Last 30 Days

Find the users with the most orders in the last 30 days of the data, counting back from the most recent order date.

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 the users with the most orders in the last 30 days of the data, counting back from the most recent order date.

A correct answer returns 6 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
1491
961
691
511
381
301

The complete result: 6 rows. This is the output of the model solution on the practice database.

What you practise

  • Filtering rows with a WHERE condition
  • 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
  • Using the result of one query inside another
  • 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

Anchor the date filter on MAX(order_date), then GROUP BY user

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.
  2. Filter in WHERE: keep the rows where order_date is at least DATE((SELECT MAX(order_date) FROM orders), '-30 day').
  3. The subquery runs once and returns a single value. It works like this:
    1. Read from the orders table.
    2. Aggregate over all the rows that are left: compute MAX(order_date). With no GROUP BY, the result is a single row.
  4. Group the rows by user_id, so the result has one row for each value.
  5. For each group, compute COUNT(*).
  6. Sort the result by cnt descending.

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.
  • Date functions. These date functions are SQLite syntax. MySQL uses DATE_FORMAT and DATEDIFF, PostgreSQL uses TO_CHAR, EXTRACT and date subtraction.
  • Subquery. A subquery compared with =, > or < must return exactly one row and one column. If it can return several rows, use IN or an aggregate.
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