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๋ฐ์ํ
https://school.programmers.co.kr/learn/courses/30/lessons/131532
ํ๋ก๊ทธ๋๋จธ์ค
SW๊ฐ๋ฐ์๋ฅผ ์ํ ํ๊ฐ, ๊ต์ก์ Total Solution์ ์ ๊ณตํ๋ ๊ฐ๋ฐ์ ์ฑ์ฅ์ ์ํ ๋ฒ ์ด์ค์บ ํ
programmers.co.kr
ํ๋ก๊ทธ๋๋จธ์ค ๋ ๋ฒจ 4
mysql group by ๋ฌธ์ ์ด๋ค.
๊ตฌ๋งคํ ์ ์ ์๋ฅผ ์นด์ดํ ํ๋ฏ๋ก distinct๋ฅผ ๊ผญ ์ฌ์ฉํด์ค์ผ ํ๋ค!!!!
๋์ ํ์ด
select year(SALES_DATE) as YEAR, month(SALES_DATE) as MONTH, GENDER, count(distinct(a.USER_ID)) as USERS
from USER_INFO a, ONLINE_SALE b
where a.USER_ID = b.USER_ID and a.GENDER is not null
group by YEAR, MONTH, GENDER
order by YEAR, MONTH, GENDER;

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