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[solvesql] 난이도 3, 4 문제 #52
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가구 판매의 비중이 높았던 날 찾기소수점 처리를 위해 SELECT
order_date,
COUNT(
DISTINCT (
CASE
WHEN category = 'Furniture' THEN order_id
END
)
) AS 'furniture',
ROUND(
COUNT(
DISTINCT (
CASE
WHEN category = 'Furniture' THEN order_id
END
)
) * 100.0 / COUNT(DISTINCT order_id),
2
) AS 'furniture_pct'
FROM
records
GROUP BY
order_date
HAVING
COUNT(DISTINCT order_id) >= 10
AND furniture_pct >= 40
ORDER BY
furniture_pct DESC,
order_date |
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solvesql 난이도 3 문제 풀이 입니다.
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