MediumSQL challenge #160DATEGROUP BYSUBQUERY

Count Orders per User in the Latest Year of Data

Return each user with the number of orders they placed in the most recent year that has orders.

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

Count orders per user in the most recent year that has orders.

A correct answer returns 56 rows with 2 columns: user_id, orders.

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
42
51
82
101
111

First 5 of 56 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 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

Hint and approach

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

Show the hint

Compare the order year with the year of MAX(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. Filter in WHERE: keep the rows where the year of order_date equals the value returned by a subquery.
  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(*).

Model solution

Checking your session…

What to watch for

The year is not typed into the query: a subquery takes the year of the latest order, so the result moves forward when newer data arrives. Both sides of the comparison come from strftime('%Y', …), so both are text and compare cleanly.

  • 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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