N042-M4 Tier 4 · Advanced · medium ecommerce · Brightlane

Return every order's ID, customer ID, order amount, and the order amount from two purchases later for that customer

Part of LAG and LEAD in SQL

The problem

Brightlane's demand forecasting team is studying longer-horizon spending trajectories. Every order should appear alongside the order amount the same customer placed two purchases later — not just the very next one.

Write a query to return every order's ID, customer ID, order amount, and the order amount from two purchases later for that customer.

Assumptions:

  • The order from two purchases later is the order whose ordered_at is the second-soonest after the current row's ordered_at, restricted to that customer.
  • For each customer's last two orders — where no second-future order is on record — the two-ahead amount is missing.
  • 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, and order_2_ahead. 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,
  LEAD(total_amount, 2) OVER (
    PARTITION BY
      customer_id
    ORDER BY
      ordered_at
  ) AS order_2_ahead
FROM
  orders
ORDER BY
  customer_id,
  ordered_at

The shape

The second argument to LEAD is an offset: LEAD(total_amount, 2) reaches two rows forward inside the partition's order rather than the default one row. Every order ends up next to the amount the same customer placed two orders later, skipping over the immediately-following order entirely.

Clause by clause

  • SELECT id, customer_id, total_amount, LEAD(total_amount, 2) OVER (PARTITION BY customer_id ORDER BY ordered_at) AS order_2_ahead returns each order's identifying columns and the amount of the customer's order two purchases ahead. PARTITION BY customer_id confines the lookup to one customer at a time; ORDER BY ordered_at sets the chronological direction; LEAD(..., 2) advances exactly two positions inside that sequence.
  • FROM orders reads every order.
  • ORDER BY customer_id, ordered_at sorts the result for chronological reading.

Why the offset and not chaining LEAD

LEAD(total_amount, 2) is a single window-function call; doing this without the offset would require nesting one LEAD inside another, which is not how window functions compose. The offset argument is the supported way to reach an arbitrary fixed distance ahead. The same applies to LAG, which accepts the offset in the same position.

The trap

A customer with only one or two recorded orders has no row that sits two positions ahead, so LEAD(..., 2) returns NULL on those rows. The number of NULLs at the tail of each customer's history equals the offset. On a customer with three orders, the last two rows return NULL for order_2_ahead; on a customer with two orders, both rows return NULL. The NULL is correct; there is no two-ahead order to read. But a downstream calculation that expects every row to carry a numeric value will silently lose rows or propagate NULLs through arithmetic.

You practiced LEAD(column, 2) — the second argument sets the offset distance; 2 jumps forward two records in the ordered partition rather than the default one.

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