๐Ÿ’ SQL/Join

[ํ”„๋กœ๊ทธ๋ž˜๋จธ์Šค] SQL ๊ณ ๋“์  Kit : ์กฐ๊ฑด์— ๋งž๋Š” ๋„์„œ์™€ ์ €์ž ๋ฆฌ์ŠคํŠธ ์ถœ๋ ฅํ•˜๊ธฐ

์„ ๋‹ฌ 2023. 4. 27. 23:19
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https://school.programmers.co.kr/learn/courses/30/lessons/144854

 

ํ”„๋กœ๊ทธ๋ž˜๋จธ์Šค

์ฝ”๋“œ ์ค‘์‹ฌ์˜ ๊ฐœ๋ฐœ์ž ์ฑ„์šฉ. ์Šคํƒ ๊ธฐ๋ฐ˜์˜ ํฌ์ง€์…˜ ๋งค์นญ. ํ”„๋กœ๊ทธ๋ž˜๋จธ์Šค์˜ ๊ฐœ๋ฐœ์ž ๋งž์ถคํ˜• ํ”„๋กœํ•„์„ ๋“ฑ๋กํ•˜๊ณ , ๋‚˜์™€ ๊ธฐ์ˆ  ๊ถํ•ฉ์ด ์ž˜ ๋งž๋Š” ๊ธฐ์—…๋“ค์„ ๋งค์นญ ๋ฐ›์œผ์„ธ์š”.

programmers.co.kr

 

-- ์ฝ”๋“œ๋ฅผ ์ž…๋ ฅํ•˜์„ธ์š”
SELECT BOOK_ID, AUTHOR_NAME, date_format(PUBLISHED_DATE, "%Y-%m-%d") as PUBLISHED_DATE
from BOOK b join AUTHOR a on b.AUTHOR_ID = a.AUTHOR_ID
where CATEGORY  = '๊ฒฝ์ œ'
order by PUBLISHED_DATE

 

 

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