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yeonkΒ·2022λ…„ 3μ›” 4일
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πŸ’‘ MySQL






πŸ”— 문제

μž…μ–‘ μ‹œκ° κ΅¬ν•˜κΈ°(1)
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πŸ’» μ½”λ“œ

SELECT HOUR(DATETIME) AS HOUR, COUNT(DATETIME) AS COUNT
FROM ANIMAL_OUTS
GROUP BY HOUR(DATETIME)
HAVING  HOUR >= 9 AND HOUR <=19
ORDER BY HOUR ASC;






참고 자료

[ν”„λ‘œκ·Έλž˜λ¨ΈμŠ€] μž…μ–‘ μ‹œκ° κ΅¬ν•˜κΈ°(1), (2) (GROUP BY, HAVING, SET)

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