HardSQL challenge #326GROUP BYDISTINCTANALYSIS

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Find users with most distinct order days.

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 most distinct order days.

A correct answer returns 114 rows with 2 columns: user_id, active_days.

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_idactive_days
775
375
1244
1184
834

First 5 of 114 rows. This is the output of the model solution on the practice database.

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
  • Sorting the result with ORDER BY

Hint and approach

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

Show the hint

COUNT DISTINCT order_date

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. Group the rows by user_id, so the result has one row for each value.
  3. For each group, compute COUNT(DISTINCT order_date), which counts each different value once.
  4. Sort the result by active_days descending.

Model solution

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What to watch for

  • 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.
  • ORDER BY. Without ORDER BY the row order is not guaranteed, even if it looks stable. Rows that tie on the sort columns can come back in any order.
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