N049-M2 Tier 4 · Advanced · medium ecommerce · Brightlane

Return every category ID, the total number of products in that category, the count of products with `price` greater than `$100`, and the average `price` among products with `price` greater than `$100`

Part of FILTER Clause on Aggregates in SQL

The problem

Brightlane's merchandising team is analyzing price distribution across product categories.

Write a query to return every category ID, the total number of products in that category, the count of products with price greater than $100, and the average price among products with price greater than $100.

Assumptions:

  • The products table has one row per product with a category_id and a price.
  • Each category_id with at least one product should appear once.
  • For each category, the total count covers every product. The above-$100 count and the above-$100 average cover only products whose price is strictly greater than $100.

Output:

  • One row per category, with columns category_id, total_products, above_100_count, and avg_above_100_price.
Schema · ecommerce 5 tables
categories
id integer
name text
parent_id? integer
products
id integer
name text
category_id integer
price numeric
stock_qty integer
attributes? jsonb
order_items
id integer
order_id integer
product_id integer
quantity integer
unit_price numeric
customers
id integer
name text
email text
city? text
country text
created_at timestamptz
is_active boolean
orders
id integer
customer_id integer
ordered_at timestamptz
status text
total_amount numeric

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Solution query
SELECT
  category_id,
  COUNT(*) AS total_products,
  COUNT(*) FILTER (
    WHERE
      price > 100
  ) AS above_100_count,
  AVG(price) FILTER (
    WHERE
      price > 100
  ) AS avg_above_100_price
FROM
  products
GROUP BY
  category_id

The shape

Two aggregates share the same FILTER (WHERE price > 100): one counts the above-$100 products and the other averages their prices. Both are restricted to the same subset of each category_id partition, while a third aggregate — COUNT(*) — runs unfiltered across the whole partition.

Clause by clause

  • SELECT category_id, COUNT(*) AS total_products, COUNT(*) FILTER (WHERE price > 100) AS above_100_count, AVG(price) FILTER (WHERE price > 100) AS avg_above_100_price returns each category's full product count, the count of products above $100, and the average price among that above-$100 subset. The two FILTER clauses target the same condition but feed different aggregates, so COUNT and AVG describe the same restricted set with different statistics.
  • FROM products reads the product records.
  • GROUP BY category_id partitions the rows per category, including the partition where category_id is NULL — GROUP BY treats every NULL as a single group, so the three products with no category_id collapse into one row.

The trap

AVG(price) FILTER (WHERE price > 100) is not the same as AVG(price) > 100. The first computes the average across only those products whose price exceeds $100, ignoring the cheaper ones entirely. A category like category 8 with five products and one above $100 (priced at $129.99) shows an above-$100 average of 129.99 — that single product's price, not a category-wide average. For a category with zero above-$100 products, both the FILTER-ed AVG and SUM return NULL because the aggregate sees an empty input. COUNT(*) FILTER returns 0 in the same situation; counts and sums diverge on empty inputs.

You practiced two parallel FILTER aggregates sharing the same condition — COUNT and AVG both restricted to the same subset, computing different statistics within that subset.

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