HardSQL challenge #336GROUP BYAGGREGATIONANALYSIS

Highest Avg Sales per Order

Find products with highest average quantity per order.

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

Solve This SQL Challenge

The task

Find products with highest average quantity per order.

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

Tables used in this challenge

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
idorder_idproduct_idquantityunit_pricesubtotal
11765192.21961.05
21823107.1321.3
316515.395.39

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_idavg_qty
495
455
704.333333333333333
1194.285714285714286
1054.2

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
  • 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 order_id)

Show the step-by-step approach

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

  1. Read from the order_items table.
  2. Group the rows by product_id, so the result has one row for each value.
  3. For each group, compute SUM(quantity) and COUNT(DISTINCT order_id), which counts each different value once.
  4. Calculate avg_qty from those aggregates: SUM(quantity) * 1.0 / COUNT(DISTINCT order_id).
  5. Sort the result by avg_qty descending.

Model solution

Checking your session…

What to watch for

  • 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.
  • Calculated columns. Dividing one integer by another truncates the result in SQLite. Multiply one side by 1.0 to get decimals.
Solve This SQL Challenge

Keep learning