HardSQL challenge #385JOINGROUP BYANALYSIS

Sales per Active Day Products

Find products with highest average sales per active day.

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

Solve This SQL Challenge

The task

Find products with highest average sales per active day.

A correct answer returns 119 rows with 2 columns: product_id, avg_sales.

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

`order_items` 496 rows

id INTEGER (PK) · order_id INTEGER · product_id INTEGER · quantity INTEGER · unit_price REAL · subtotal REAL

First 3 rows of the order_items table
id	order_id	product_id	quantity	unit_price	subtotal
1	1	76	5	192.21	961.05
2	1	82	3	107.1	321.3
3	1	65	1	5.39	5.39

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
product_id	avg_sales
49	5
45	5
3	4.4
70	4.333333333333333
119	4.285714285714286

First 5 of 119 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
Counting different values with COUNT(DISTINCT …)
Grouping rows with GROUP BY to get one result row per group
Combining rows from two tables on a matching key with an inner join
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

SUM(quantity)/COUNT(DISTINCT day)

Show the step-by-step approach

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

Read from the order_items table.
Join orders on order_items.order_id = orders.id.
Group the rows by order_items.product_id, so the result has one row for each value.
For each group, compute SUM(order_items.quantity) and COUNT(DISTINCT orders.order_date), which counts each different value once.
Calculate avg_sales from those aggregates: SUM(order_items.quantity) * 1.0 / COUNT(DISTINCT orders.order_date).
Sort the result by avg_sales descending.

Model solution

Checking your session…

What to watch for

INNER JOIN. An inner join drops rows that have no partner on the other side, and it repeats a row once for every match, which inflates SUM and COUNT.
COUNT(DISTINCT). After a join, COUNT(*) counts joined rows. Counting the parent entity needs COUNT(DISTINCT its id).
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.