HardSQL challenge #236WINDOW FUNCTIONLAGAGGREGATION

Calculate the Day-to-Day Change in Revenue

Return each order date with the change in total revenue from the previous date.

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 change in revenue day to day.

A correct answer returns 182 rows with 2 columns: order_date, diff.

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	diff
2022-01-06	NULL
2022-01-10	2076.95
2022-01-15	-2768.31
2022-01-16	2239.5299999999997
2022-01-17	772.1100000000006

First 5 of 182 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
Reading the previous row's value with LAG()
Deriving a new column from an expression and naming it with an alias

Hint and approach

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

Show the hint

LAG SUM

Show the step-by-step approach

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

Read from the orders table.
Group the rows by order_date, so the result has one row for each value.
For each group, compute SUM(total).
Fetch SUM(total) from the previous row with LAG(), ordered by order_date. Where there is no such row the value is NULL.
Combine those values into diff: SUM(total) - LAG(SUM(total)) OVER (ORDER BY order_date).

Model solution

Checking your session…

What to watch for

GROUP BY builds the daily totals and LAG() then reads the previous day's total in the same query. The previous row is the previous date that has orders, which may be several calendar days earlier.

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.
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.
Calculated columns. Dividing one integer by another truncates the result in SQLite. Multiply one side by 1.0 to get decimals.

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