N056-E2 Tier 4 · Advanced · easy ecommerce · Brightlane

Return each calendar month, the number of `orders` placed in that month, and the number of `orders` placed in the immediately following calendar month

Part of Period-over-Period Analysis in SQL

The problem

Scenario: Brightlane's growth team is planning headcount for the coming month and wants each month's order count shown alongside the count from the following month.

Task: Write a query to return each calendar month, the number of orders placed in that month, and the number of orders placed in the immediately following calendar month.

Assumptions:

  • A calendar month is identified by its first day and covers every order placed within that month.
  • The latest month in the data has no following month; its next_month_count value is missing.

Output:

  • One row per calendar month present in the data.
  • Columns in this order: month (the first day of the calendar month), order_count, next_month_count.
  • Sorted by month ascending.
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

Run previews · Check grades

Write a query, then run it to see results here.

Worked solution Try it yourself first
Solution query
SELECT
  DATE_TRUNC('month', ordered_at)::date AS MONTH,
  COUNT(*) AS order_count,
  LEAD(COUNT(*)) OVER (
    ORDER BY
      DATE_TRUNC('month', ordered_at)
  ) AS next_month_count
FROM
  orders
GROUP BY
  DATE_TRUNC('month', ordered_at)
ORDER BY
  MONTH

The shape

LEAD is the forward-looking sibling of LAG: it reaches one row ahead in the ordered sequence instead of one row back. Each month's row carries the count from the month that follows it, which is exactly the look-ahead the headcount plan needs.

Clause by clause

  • SELECT DATE_TRUNC('month', ordered_at)::date AS month, COUNT(*) AS order_count, LEAD(COUNT(*)) OVER (ORDER BY DATE_TRUNC('month', ordered_at)) AS next_month_count returns the month bucket, the month's order count, and the following month's order count. COUNT(*) counts every row in the group; LEAD with no explicit offset advances exactly one row in the ordered sequence.
  • FROM orders reads every order in the table.
  • GROUP BY DATE_TRUNC('month', ordered_at) collapses the orders into one row per calendar month so COUNT(*) produces the monthly total.
  • ORDER BY month prints the months in calendar order so the next-month column lines up with the row it describes.

Why LEAD and not LAG

LAG would have the prior month sitting next to each current month, which is the wrong direction for a forward-looking report. LEAD reaches in the same direction the planner is reading: this month is here, and the count next month is the value to the right. Switching to LAG would silently produce the wrong column for the question being asked, with no error to flag the mistake.

The trap

The last month in the data has no following row in the window, so next_month_count is NULL for that final row. That NULL is honest: there is no measurement for the month after the last one. It is not a candidate for a default substitution unless the report's downstream consumer specifically needs a number instead of a missing value.

You practiced using LEAD over monthly order totals to attach each month's following-month value alongside the current value as an inline column.

How you actually get good at SQL

Reading explains SQL. Writing it, over and over with instant feedback, is what makes you fluent.

That's the whole SQLMaxx loop: 600+ real problems, instant AI feedback, mastery you can actually see, and spaced review that won't let you forget.

A stack of SQL practice problem cards, the top card showing an employees table.
615 problems · 66 concepts

Real problems. Not toy examples.

615 hand-built problems spanning all 66 concepts, from basic SELECTs to window functions, built on real schemas and real business questions, the kind you'll actually get asked on the job. Enough reps to make SQL automatic.

A retro computer showing a SQL query marked correct with a green checkmark.
Instant AI feedback

Write a query. Know if it's right in one second.

No copying an answer and hoping it clicked. The AI grader checks your real query against real data, catches exactly what's wrong, and explains the fix in plain English, like a senior analyst reading over your shoulder on every problem.

A circular mastery progress dial filling from blue to green, the SQLMaxx diamond at its center.
Mastery tracking

Stop guessing whether you actually know it.

SQLMaxx tracks every concept and shows you what you've mastered and what's still shaky. Your skills fill in one concept at a time, so 'I think I get joins' becomes something you can prove.

A SQL query editor circled by a blue return arrow with a clock, scheduled to come back for review.
Spaced review

Learn it once. Keep it for good.

Most of what you learn this week fades by next week. So when a concept comes due for review, SQLMaxx hands you a fresh problem to solve from a blank editor, not a flashcard to re-read. A research-backed spaced-repetition algorithm (FSRS) times each return for right before you'd forget, so your SQL is still there months later, when the interview or the job actually needs it.

Practice, feedback, mastery, review. That's the loop that turns reading into real skill.

Start free

No account, no credit card. Start solving in under a minute.