MediumSQL challenge #184JOINGROUP BYSUBQUERY

Show Review Count and Units Sold per Product

Return every product with its number of reviews and its total quantity sold.

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

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The task

For every product, show its number of reviews and its total quantity sold.

A correct answer returns 120 rows with 3 columns: id, reviews, sales.

Tables used in this challenge

products 120 rows

id INTEGER (PK) · name TEXT · category TEXT · price REAL · cost REAL · stock INTEGER · sku TEXT · status TEXT · rating REAL · review_count INTEGER

First 3 rows of the products table
idnamecategorypricecoststockskustatusratingreview_count
1Play Kitchen DeluxeToys1786.38589.51257SKU-000001active457
2Micro-Cut Paper Shredder 8-SheetOffice Supplies907.93281.46131SKU-000002active2.5447
34K Dual Dash CameraAutomotive568.89204.8176SKU-000003active4.129

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

reviews 180 rows

id INTEGER (PK) · product_id INTEGER · user_id INTEGER · rating INTEGER · comment TEXT · reviewed_at TEXT · helpful_votes INTEGER

First 3 rows of the reviews table
idproduct_iduser_idratingcommentreviewed_athelpful_votes
123455Works perfectly. Very happy with this purchase.2022-09-0240
2101685Excellent build quality. Will buy again!2023-01-0596
340391Would give 6 stars if I could!2023-05-1517

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
idreviewssales
124
239
3122
414
5122

First 5 of 120 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
  • Keeping every row of the left table with LEFT JOIN, matched or not
  • Querying the result of another query with a subquery in FROM
  • Replacing NULL with a default using COALESCE

Hint and approach

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

Show the hint

Aggregate reviews and order_items separately, then LEFT JOIN both results to products

Show the step-by-step approach

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

  1. Read from the products table.
  2. Build a second inner query (r) to join to:
    1. Read from the reviews table.
    2. Group the rows by product_id, so the result has one row for each value.
    3. For each group, compute COUNT(*).
  3. LEFT JOIN that inner query on r.product_id = products.id. Rows from the left side that have no match are kept, with NULL in the joined columns.
  4. Build a second inner query (s) to join to:
    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).
  5. LEFT JOIN that inner query on s.product_id = products.id. Rows from the left side that have no match are kept, with NULL in the joined columns.
  6. COALESCE(x, 0) replaces a NULL result with 0.

Model solution

Checking your session…

What to watch for

Each child table is reduced to one row per product before the join, so nothing can multiply. COUNT(DISTINCT) can rescue a count after a double join, but there is no equivalent rescue for SUM, which is why the pre-aggregated form is the safe default. COALESCE turns the NULL of a product with no reviews or no sales into 0.

  • Derived table. Columns computed in the inner query are only visible outside under their alias, so every calculated column there needs a name.
  • LEFT JOIN. A WHERE condition on the right-hand table removes the unmatched rows and turns the LEFT JOIN back into an inner join. Conditions on that table go in the ON clause.
  • 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.
Solve This SQL Challenge

Same technique, other tables

More practice on this pattern with different columns:

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