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N039-M1 Tier 3 · Intermediate · medium ecommerce · Brightlane

Return the ID, customer ID, and total amount of every order, plus a sequential number within that customer's orders, ordered by `total_amount` in descending order

Part of ROW_NUMBER, RANK, DENSE_RANK in SQL

The problem

Brightlane's sales analyst wants a numbered sequence of each customer's orders sorted from most expensive to least expensive.

Write a query to return the ID, customer ID, and total amount of every order, plus a sequential number within that customer's orders, ordered by total_amount in descending order.

Assumptions:

  • The orders table has one row per order with an id, a customer_id, and a total_amount.
  • Within each customer's orders, the position starts at 1 for the highest total_amount and increments by 1 for each subsequent order. Numbering restarts for each customer.
  • If two orders for the same customer share the same total_amount, they receive consecutive numbers in some order.

Output:

  • One row per order, with columns id, customer_id, total_amount, and order_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,
  customer_id,
  total_amount,
  ROW_NUMBER() OVER (
    PARTITION BY
      customer_id
    ORDER BY
      total_amount DESC
  ) AS order_rank
FROM
  orders

The shape

ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY total_amount DESC) restarts the numbering at 1 for each customer and counts down their orders from largest to smallest. The customer's biggest order is 1, second-biggest is 2, and the sequence is independent of how every other customer's orders fall.

Clause by clause

  • SELECT id, customer_id, total_amount, ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY total_amount DESC) AS order_rank returns each order's identifying columns and its rank within that customer. PARTITION BY customer_id divides the orders into one bucket per customer; ORDER BY total_amount DESC sorts within each bucket; ROW_NUMBER then numbers the rows of each bucket starting from 1.
  • FROM orders reads every order in the system. No filter is applied; every customer's full history is in the result.

Why PARTITION BY and not a global ROW_NUMBER followed by a filter

A global ROW_NUMBER() OVER (ORDER BY total_amount DESC) would return the absolute rank across the entire orders table, which is a different question. The brief asks for per-customer ranks: each customer's biggest order should be 1, not the one customer who happens to have the largest dollar order overall. PARTITION BY is the construct that resets the window between customers; without it, the rank is computed against every order in the table at once.

The trap

Ties between two orders with the same total inside one customer's history are broken arbitrarily by ROW_NUMBER. If two orders for the same customer share a total_amount, one will get 1 and the other 2 even though the data says nothing about which is "first." Adding a secondary sort like ORDER BY total_amount DESC, id makes the tie-break deterministic. RANK or DENSE_RANK would carry the tie into the output instead of hiding it.

You practiced ROW_NUMBER() OVER (PARTITION BY ... ORDER BY ...) — partition the numbering so each partition restarts at 1 independently.

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