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N039-M2 Tier 3 · Intermediate · medium ecommerce · Brightlane

Return the ID, name, category ID, and price of every product, plus the product's rank within its category by `price` in descending order

Part of ROW_NUMBER, RANK, DENSE_RANK in SQL

The problem

Brightlane's category manager ranks products within each category by price from highest to lowest.

Write a query to return the ID, name, category ID, and price of every product, plus the product's rank within its category by price in descending order.

Assumptions:

  • The products table has one row per product with an id, a name, a category_id, and a price.
  • Within each category, rank 1 goes to the highest-priced product. Rank values restart for each category.
  • Products with the same price in the same category receive the same rank, with the next rank adjusted upward by the number of tied products.

Output:

  • One row per product, with columns id, name, category_id, price, and category_price_rank.
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
  id,
  name,
  category_id,
  price,
  RANK() OVER (
    PARTITION BY
      category_id
    ORDER BY
      price DESC
  ) AS category_price_rank
FROM
  products

The shape

RANK() OVER (PARTITION BY category_id ORDER BY price DESC) runs an independent price ranking inside each category. The most expensive product in a category gets 1 no matter what the global top price is, and ties on price within a category produce shared ranks followed by a gap.

Clause by clause

  • SELECT id, name, category_id, price, RANK() OVER (PARTITION BY category_id ORDER BY price DESC) AS category_price_rank returns each product's identifying columns and its rank inside its category. PARTITION BY category_id separates the catalog into one window per category; ORDER BY price DESC sorts within each category from most to least expensive; RANK assigns rank 1 to the top product (or products) in each category.
  • FROM products reads every product. No filter; every product receives a per-category rank.

Why RANK and not DENSE_RANK

DENSE_RANK would also produce per-category rankings, but it would not gap after a tie. If a category has two products tied at the top price, RANK returns 1, 1, 3, ... and DENSE_RANK returns 1, 1, 2, .... The brief asks for a "price rank" where the gap is informative: a product at rank 5 in its category has four strictly-pricier products above it. DENSE_RANK would obscure that signal by collapsing tied positions.

The trap

The rank resets at every category_id boundary, which means a product showing category_price_rank = 1 is not the most expensive in the catalog; it is the most expensive in its category. Reading the output without keeping the partition column in mind is the easy mistake. Always carry category_id (or whatever column the partition is on) in the SELECT list so the rank is interpretable.

You practiced RANK() OVER (PARTITION BY ... ORDER BY ...) — tie-aware ranking that restarts within each partition; gaps appear after every tie.

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