๐ ์๊ณ ๋ฆฌ์ฆ/Programmers - SQL
[MySQL/PGS] Lv.3 : ๋์ฅ๊ท ๋ค์ ์์์ ์ ๊ตฌํ๊ธฐ
xxilliant
2025. 5. 7. 15:37
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๋ฐ์ํ
https://school.programmers.co.kr/learn/courses/30/lessons/299305
ํ๋ก๊ทธ๋๋จธ์ค
SW๊ฐ๋ฐ์๋ฅผ ์ํ ํ๊ฐ, ๊ต์ก, ์ฑ์ฉ๊น์ง Total Solution์ ์ ๊ณตํ๋ ๊ฐ๋ฐ์ ์ฑ์ฅ์ ์ํ ๋ฒ ์ด์ค์บ ํ
programmers.co.kr
ํ๋ก๊ทธ๋๋จธ์ค ๋ ๋ฒจ 3.
์ค๋๋ง์ sql ํธ๋๊น ์๊ฐ๋ณด๋ค ์ด๋ ต๋ค..ใ .ใ
mysql left join, if ~ is null, ์๋ธ์ฟผ๋ฆฌ ๋ฑ์ ์ฌ์ฉํ์ฌ ํด๊ฒฐํ๋ค.
๋์ ํ์ด
select a.ID,
if(b.CHILD_COUNT is null, 0, b.CHILD_COUNT) as CHILD_COUNT
from ecoli_data a
left join (select PARENT_ID, count(*) as CHILD_COUNT
from ecoli_data
where PARENT_ID is not null
group by parent_id) b on a.ID = b.PARENT_ID;
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๋ฐ์ํ