HardSQL challenge #380GROUP BYDISTINCTANALYSIS

Orders per Active Day

Find users with highest ratio of total orders to active 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 highest ratio of total orders to active days.

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

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_idorders_per_day
1501
1491
1481
1471
1461

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
  • Deriving a new column from an expression and naming it with an alias

Hint and approach

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

Show the hint

COUNT orders / 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(*) and COUNT(DISTINCT order_date), which counts each different value once.
  4. Calculate orders_per_day from those aggregates: COUNT(*) * 1.0 / COUNT(DISTINCT order_date).
  5. Sort the result by orders_per_day 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.
  • Calculated columns. Dividing one integer by another truncates the result in SQLite. Multiply one side by 1.0 to get decimals.
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