MediumSQL challenge #112DATESUBQUERY

Find Orders from the Last 30 Days of Data

Return the orders placed in the 30 days up to the most recent order date in the table.

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 orders placed in the last 30 days of the data, counting back from the most recent order date.

A correct answer returns 6 rows with 10 columns: id, user_id, total, discount, item_count, status, order_date, shipped_date, shipping_city, shipping_country.

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
iduser_idtotaldiscountitem_countstatusorder_dateshipped_dateshipping_cityshipping_country
701492104.220.173pending2024-12-24NULLPhiladelphiaUSA
97514566.230.161delivered2024-12-092023-01-27Los AngelesUSA
12830370.050.135refunded2024-12-13NULLNew YorkUSA
15569110.60.116cancelled2024-12-19NULLSan JoseUSA
169961308.160.095delivered2024-12-122023-04-18JacksonvilleUSA
171382162.10.044delivered2024-12-232023-02-08IndianapolisUSA

The complete result: 6 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

  • Returning every column of a table with SELECT *
  • Filtering rows with a WHERE condition
  • Summarising many rows into one value with COUNT, SUM, AVG, MIN or MAX
  • 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

Anchor on MAX(order_date) instead of today's date, then use DATE(..., '-30 day')

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 order_date is at least DATE((SELECT MAX(order_date) FROM orders), '-30 day').
  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. Return every column with SELECT *.

Model solution

Checking your session…

What to watch for

The usual DATE('now', '-30 day') returns nothing on a dataset that ends in the past, as this one does (December 2024). Anchoring on MAX(order_date) makes the window follow the data. On a live table you would use 'now' instead.

  • 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.
  • 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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Same technique, other tables

More practice on this pattern with different columns:

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