[LeetCode] 1211. Queries Quality and Percentage

 

 

[문제]

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[MySQL]

WITH evaluation AS (
    SELECT *
        , rating / position AS quality
        , CASE WHEN rating < 3 THEN 'poor' END AS poor_query
    FROM Queries
)

SELECT query_name
     , ROUND(SUM(quality) / COUNT(*), 2) AS quality
     , ROUND(COUNT(poor_query) / COUNT(*) * 100, 2) AS poor_query_percentage
FROM evaluation
WHERE query_name IS NOT NULL
GROUP BY query_name