HardSQL challenge #290WINDOW FUNCTIONLAGPATTERN

Find Days Where Revenue Rose Twice in a Row

Return the order dates on which daily revenue was above the previous day, which was above the day before.

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.

Solve This SQL Challenge

The task

Find the order dates on which daily revenue rose for the second time in a row: higher than the previous order day, which was itself higher than the one before.

A correct answer returns 29 rows with one column: order_date.

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
id	user_id	total	discount	item_count	status	order_date	shipped_date	shipping_city	shipping_country
1	43	2638.11	0.25	3	cancelled	2024-04-24	NULL	Los Angeles	USA
2	8	2947	0.19	5	delivered	2024-03-15	2023-04-09	Chicago	USA
3	136	683.24	0.08	2	delivered	2023-02-15	2023-07-10	Los Angeles	USA

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
order_date
2022-01-17
2022-03-03
2022-05-16
2022-05-26
2022-06-17

First 5 of 29 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

Choosing exactly which columns a query returns
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
Querying the result of another query with a subquery in FROM
Reading the previous row's value with LAG()

Hint and approach

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

Show the hint

Total revenue per day first, then use LAG with offsets 1 and 2

Show the step-by-step approach

How the model solution is built, in the order the database evaluates it:

Build an inner query first (a derived table):
1. Build an inner query first (a derived table):
  1. Read from the orders table.
  2. Group the rows by order_date, so the result has one row for each value.
  3. For each group, compute SUM(total).
2. Fetch revenue from the previous row with LAG(), ordered by order_date. Where there is no such row the value is NULL.
3. Fetch revenue from 2 rows back with LAG(), ordered by order_date. Where there is no such row the value is NULL.
Filter in WHERE: keep the rows where revenue is greater than prev1, and prev1 is greater than prev2. The filter sits in the outer query because a window function cannot be tested in the WHERE clause of the query that computes it.
Return only the column order_date.

Model solution

Checking your session…

What to watch for

LAG takes an offset as its second argument: LAG(revenue, 2) reads two rows back. Revenue is totalled per day first, because comparing raw order rows would treat several orders on one day as separate days. The first two dates have NULL comparisons and can never qualify.

LAG(). LAG() returns NULL for the first row of each partition, and any arithmetic with that NULL is NULL too. Supply a default as the third argument if you need a number.
Derived table. Columns computed in the inner query are only visible outside under their alias, so every calculated column there needs a name.
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.