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Given a table with sales data, write a query to find consecutive days with decreasing revenue.

SQLmedium0.3 min read

Reviewed by Aditya Kumar · Last reviewed 2026-03-24

**LAG approach**: WITH t AS (SELECT date, revenue, LAG(revenue) OVER (ORDER BY date) AS prev_rev FROM sales) SELECT date, revenue FROM t WHERE revenue < prev_rev AND prev_rev IS NOT NULL. **Consecutive pairs**: Self-join ON a.date = b.date + 1 WHERE a.revenue < b.revenue. **N...

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Asked at 1 company
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63 words

LAG approach: WITH t AS (SELECT date, revenue, LAG(revenue) OVER (ORDER BY date) AS prev_rev FROM sales) SELECT date, revenue FROM t WHERE revenue < prev_rev AND prev_rev IS NOT NULL. Consecutive pairs: Self-join ON a.date = b.date + 1 WHERE a.revenue < b.revenue. N consecutive: Extend with multiple LAG or recursive pattern. Gaps: Date sequence must be continuous or use date diff.

⚡
Pro Tip

Red Flag: Ignoring date gaps (weekends, holidays). Pro-Move: 'We used ROW_NUMBER() OVER (ORDER BY date) - date diff to group consecutive days, then LAG for trend—handled gaps correctly.'

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