MediumSQL challenge #150JOINGROUP BY

Count Orders and Total Spend for Every User, Including Zero

Return every user with their number of orders and total spend, including users with none.

The challenge runs on the SQLabHub practice database (SQLite) and uses the users and orders tables. You can solve it in the browser, with nothing to install.

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The task

Find total orders and total spend per user.

A correct answer returns 150 rows with 3 columns: id, total_orders, total_spend.

Tables used in this challenge

users 150 rows

id INTEGER (PK) · first_name TEXT · last_name TEXT · email TEXT · phone TEXT · city TEXT · country TEXT · created_at TEXT · status TEXT · is_premium INTEGER

First 3 rows of the users table
idfirst_namelast_nameemailphonecitycountrycreated_atstatusis_premium
1JenniferTorresjennifer.torres57@icloud.com+1-263-376-2371HoustonCanada2024-04-13active0
2DonnaAdamsdonna.adams52@gmail.com+1-753-780-6670Fort WorthCanada2020-09-15inactive0
3KennethYoungkenneth.young94@proton.me+1-440-743-9503AustinUSA2023-09-05active0

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
idtotal_orderstotal_spend
114064.75
211618.29
30NULL
434863.29
514924.23

First 5 of 150 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
  • Keeping every row of the left table with LEFT JOIN, matched or not

Hint and approach

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

Show the hint

JOIN + 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 users table.
  2. LEFT JOIN orders on users.id = orders.user_id. Rows from the left side that have no match are kept, with NULL in the joined columns.
  3. Group the rows by users.id, so the result has one row for each value.
  4. For each group, compute COUNT(orders.id) and SUM(orders.total).

Model solution

Checking your session…

What to watch for

All 150 users come back. Those with no orders show a count of 0, because COUNT(o.id) skips the NULLs the LEFT JOIN produces, and a NULL total, because SUM of nothing is NULL. Wrap it in COALESCE to show 0.

  • LEFT JOIN. A WHERE condition on the right-hand table removes the unmatched rows and turns the LEFT JOIN back into an inner join. Conditions on that table go in the ON clause.
  • 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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