MediumSQL challenge #125GROUP BY

Orders by Shipping City

Count orders grouped by shipping city.

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 grouped by shipping city.

A correct answer returns 20 rows with 2 columns: shipping_city, COUNT(*).

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
shipping_cityCOUNT(*)
Austin7
Charlotte11
Chicago9
Columbus12
Dallas11

First 5 of 20 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

  • 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

Hint and approach

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

Show the hint

GROUP BY

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 shipping_city, so the result has one row for each value.
  3. For each group, compute COUNT(*).

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.
  • Aggregate functions. Aggregates skip NULLs. AVG divides by the number of non-NULL values and COUNT(column) can be smaller than COUNT(*).
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