N043-H2 Tier 4 · Advanced · hard ecommerce · Brightlane

Return every order's ID, customer ID, order amount, the amount of that same customer's first order chronologically, and the amount of that same customer's most recent order chronologically

Part of FIRST_VALUE, LAST_VALUE, NTH_VALUE in SQL

The problem

Brightlane's customer analytics team needs every order annotated with both ends of that customer's purchase history — the amount of the customer's very first order and the amount of the customer's most recent order, attached to every row.

Write a query to return every order's ID, customer ID, order amount, the amount of that same customer's first order chronologically, and the amount of that same customer's most recent order chronologically.

Assumptions:

  • A customer's first order is the order with the smallest ordered_at for that customer_id. A customer's most recent order is the order with the largest ordered_at for that customer_id.
  • Both anchor values appear identically on every row sharing a customer_id, including rows that fall in the middle of the customer's chronological sequence.
  • The final result is sorted by customer_id ascending, then by ordered_at ascending.

Output:

  • One row per order, with columns id, customer_id, total_amount, first_order_amount, and last_order_amount. Sorted by customer_id, then ordered_at.
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,
  FIRST_VALUE(total_amount) OVER (
    PARTITION BY
      customer_id
    ORDER BY
      ordered_at
  ) AS first_order_amount,
  LAST_VALUE(total_amount) OVER (
    PARTITION BY
      customer_id
    ORDER BY
      ordered_at ROWS BETWEEN UNBOUNDED PRECEDING
      AND UNBOUNDED FOLLOWING
  ) AS last_order_amount
FROM
  orders
ORDER BY
  customer_id,
  ordered_at

The shape

Two window functions partitioned identically by customer_id and ordered by ordered_at produce two anchor annotations on the same row. FIRST_VALUE works with the default frame because position 1 is in every running frame. LAST_VALUE does not, and needs an explicit ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING frame to reach the partition's true last position.

Clause by clause

SELECT id, customer_id, total_amount,
  FIRST_VALUE(total_amount) OVER (PARTITION BY customer_id ORDER BY ordered_at) AS first_order_amount,
  LAST_VALUE(total_amount) OVER (
    PARTITION BY customer_id ORDER BY ordered_at
    ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING
  ) AS last_order_amount
FROM orders
ORDER BY customer_id, ordered_at
  • The FIRST_VALUE window has no explicit frame. The default frame ends at the current row, but position 1 of the partition is inside that frame on every row from row 1 onward, so the default works.
  • The LAST_VALUE window adds ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING to extend the frame across the whole partition. Without it, LAST_VALUE would return the current row's value on every row, because the default frame's tail is the current row.
  • Both functions partition by customer_id and order by ordered_at, so they walk each customer's history in the same chronological sequence.
  • The outer ORDER BY customer_id, ordered_at sorts the printed result; this is independent of either window's ordering.

Why two separate OVER clauses and not one shared window

The two functions need different window definitions because they need different frames. FIRST_VALUE is correct under the default frame; LAST_VALUE requires the explicit full-partition frame. Writing the two OVER clauses independently makes each function's frame visible in the SQL. Forcing the two functions to share a WINDOW alias would require either widening FIRST_VALUE's frame unnecessarily or breaking LAST_VALUE by leaving the frame default.

The trap

The two windows look almost identical on the page and behave almost identically in the partition: same PARTITION BY, same ORDER BY. The frame difference is small in text and decisive in result. Copy the FIRST_VALUE window verbatim to LAST_VALUE and every row's last_order_amount will equal its own total_amount, which looks plausible for the last row of each customer and is wrong for every other row. Whenever a single query uses both FIRST_VALUE and LAST_VALUE, the rule is fixed: FIRST_VALUE may use the default frame; LAST_VALUE must use ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING.

You practiced FIRST_VALUE and LAST_VALUE over the same partition — FIRST_VALUE works correctly with the default frame; LAST_VALUE requires an explicit ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING frame so the lookup spans the whole partition rather than ending at the current row.

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