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[SQL] LEFT JOIN์ ํ์ฉํ ๋๋ฝ๋ ๋ฐ์ดํฐ ์ฒ๋ฆฌํ๊ธฐ
ํด๋ฒ๋
2024. 4. 19. 14:34
๋ฐ์ํ
์ ํ๋ฅผ ๋ณด๋ฉด codetype์ fruit์ด ์๊ณ codekey์๋ ๊ณผ์ผ์ ์ข ๋ฅ๊ฐ ์ ํ์๋ค.
๊ทธ๋ฆฌ๊ณ ๊ฐ๊ฐ ๊ณผ์ผ์ ๋ํ ์ข ๋ฅ์ ์ด๋ฆ์ด ๋์์๋ค.
ํ์ฌ ํฌ๋๋ 4๊ฐ, ์ฌ๊ณผ๋ 2๊ฐ, ์ค๋ ์ง๋ 3๊ฐ, ๋ฐฐ๋ 0๊ฐ์ธ๋ฐ
ํฌ๋, ์ฌ๊ณผ, ์ค๋ ์ง, ๋ฐฐ ๋ณ๋ก ๊ฐ์๋ฅผ ์ธ๋ ์ฟผ๋ฆฌ๋ฌธ์ ๋ง๋ค๊ณ ์ถ์๋ค.
ํญ๋ชฉ์ ์๋ ๊ฑด ์ ์ธํ๊ณ count ํ๊ธฐ
select B.codekey, count(A.codekey)
from (select * from fruit where codetype like 'fruit') B
join fruit A
on A.codetype = B.codekey
group by B.codekey;
ํญ๋ชฉ์ ์๋ ๊ฒ๋ count ํ๊ธฐ
๋ฐฐ๋ 0๊ฐ์ธ๋ฐ 0๊ฐ๋ ์ถ๋ ฅํ๊ณ ์ถ๋ค!
left join์ ์ฌ์ฉํ์ฌ ๋ง๋ค์๋ค.
select B.codekey, count(A.codekey)
from (select * from fruit where codetype like 'fruit') B
left join fruit A
on A.codetype = B.codekey
group by B.codekey;
๋ฐ์ํ