HardSQL challenge #383WINDOW FUNCTIONLAGAGGREGATION

Highest Daily Growth Orders

Find order days with highest increase in revenue compared to previous day.

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 order days with highest increase in revenue compared to previous day.

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

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
order_dategrowth
2024-04-217415.110000000001
2023-12-236305.130000000001
2023-03-186156.889999999999
2022-08-225351.5599999999995
2023-08-244692.719999999999

First 5 of 182 rows. This is the output of the model solution on the practice database.

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()
  • Sorting the result with ORDER BY
  • 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:

  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).
  4. Fetch SUM(total) from the previous row with LAG(), ordered by order_date. Where there is no such row the value is NULL.
  5. Combine those values into growth: SUM(total) - LAG(SUM(total)) OVER (ORDER BY order_date).
  6. Sort the result by growth descending.

Model solution

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What to watch for

  • 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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