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https://school.programmers.co.kr/learn/courses/30/lessons/59041
ํ๋ก๊ทธ๋๋จธ์ค
SW๊ฐ๋ฐ์๋ฅผ ์ํ ํ๊ฐ, ๊ต์ก์ Total Solution์ ์ ๊ณตํ๋ ๊ฐ๋ฐ์ ์ฑ์ฅ์ ์ํ ๋ฒ ์ด์ค์บ ํ
programmers.co.kr
ํ๋ก๊ทธ๋๋จธ์ค ๋ ๋ฒจ 2 mysql
๋์ ํ์ด
SELECT NAME, COUNT(NAME) AS COUNT
FROM ANIMAL_INS
WHERE NAME IS NOT NULL
GROUP BY NAME
HAVING COUNT > 1
ORDER BY NAME;

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