DB/SQL

SQL - HAVING

giraffe_ 2022. 6. 13. 11:55

https://programmers.co.kr/learn/courses/30/lessons/59041

 

์ฝ”๋”ฉํ…Œ์ŠคํŠธ ์—ฐ์Šต - ๋™๋ช… ๋™๋ฌผ ์ˆ˜ ์ฐพ๊ธฐ

ANIMAL_INS ํ…Œ์ด๋ธ”์€ ๋™๋ฌผ ๋ณดํ˜ธ์†Œ์— ๋“ค์–ด์˜จ ๋™๋ฌผ์˜ ์ •๋ณด๋ฅผ ๋‹ด์€ ํ…Œ์ด๋ธ”์ž…๋‹ˆ๋‹ค. ANIMAL_INS ํ…Œ์ด๋ธ” ๊ตฌ์กฐ๋Š” ๋‹ค์Œ๊ณผ ๊ฐ™์œผ๋ฉฐ, ANIMAL_ID, ANIMAL_TYPE, DATETIME, INTAKE_CONDITION, NAME, SEX_UPON_INTAKE๋Š” ๊ฐ๊ฐ ๋™๋ฌผ์˜ ์•„์ด๋””

programmers.co.kr

 

 

 

 

 

oracle

SELECT NAME, COUNT(NAME)
FROM ANIMAL_INS
GROUP BY NAME
HAVING COUNT(NAME) >= 2
ORDER BY NAME

 

 

 

 

 

https://programmers.co.kr/learn/courses/30/lessons/59412?language=mysql 

 

์ฝ”๋”ฉํ…Œ์ŠคํŠธ ์—ฐ์Šต - ์ž…์–‘ ์‹œ๊ฐ ๊ตฌํ•˜๊ธฐ(1)

ANIMAL_OUTS ํ…Œ์ด๋ธ”์€ ๋™๋ฌผ ๋ณดํ˜ธ์†Œ์—์„œ ์ž…์–‘ ๋ณด๋‚ธ ๋™๋ฌผ์˜ ์ •๋ณด๋ฅผ ๋‹ด์€ ํ…Œ์ด๋ธ”์ž…๋‹ˆ๋‹ค. ANIMAL_OUTS ํ…Œ์ด๋ธ” ๊ตฌ์กฐ๋Š” ๋‹ค์Œ๊ณผ ๊ฐ™์œผ๋ฉฐ, ANIMAL_ID, ANIMAL_TYPE, DATETIME, NAME, SEX_UPON_OUTCOME๋Š” ๊ฐ๊ฐ ๋™๋ฌผ์˜ ์•„์ด๋””, ์ƒ๋ฌผ

programmers.co.kr

 

 

 

 

 

MySQL

SELECT HOUR(DATETIME) AS HOUR, COUNT(HOUR(DATETIME)) AS COUNT
FROM ANIMAL_OUTS
GROUP BY HOUR(DATETIME)
HAVING HOUR >= 9 AND HOUR <= 19 
ORDER BY HOUR