없어진 기록 찾기 select outs.animal_id as ANIMAL_ID, outs.name as NAME from ANIMAL_OUTS as outs left outer join ANIMAL_INS as ins on outs.animal_id = ins.animal_id where ins.animal_id is null 있었는데요 없었습니다. select ANIMAL_INS.animal_id, ANIMAL_INS.name from ANIMAL_INS inner join ANIMAL_OUTS AO on ANIMAL_INS.animal_id = AO.animal_id and ANIMAL_INS.datetime > AO.datetime order by ANIMAL_INS.datetime 오랜 기간 보..

이름이 없는 동물의 아이디 select animal_id from ANIMAL_INS where name is null order by animal_id 이름이 있는 동물의 아이디 select animal_id from ANIMAL_INS where name is not null order by animal_id NULL 처리하기 select animal_type, ifnull(name, 'No name') as name ,sex_upon_intake from ANIMAL_INS order by animal_id

고양이와 개는 몇 마리 있을까 # (1) SELECT ANIMAL_TYPE as TYPE, COUNT(ANIMAL_TYPE) as COUNT FROM ANIMAL_INS WHERE ANIMAL_TYPE IN ('Dog', 'Cat') GROUP BY ANIMAL_TYPE # (2) SELECT ANIMAL_TYPE, COUNT(ANIMAL_TYPE) as count FROM ANIMAL_INS GROUP BY ANIMAL_TYPE ORDER BY ANIMAL_TYPE 동명 동물 수 찾기 SELECT t1.NAME, t1.COUNT FROM (SELECT NAME, COUNT(NAME) as COUNT FROM ANIMAL_INS WHERE NAME IS NOT NULL GROUP BY NAME) as t..

최댓값 구하기 # Using filesort SELECT DATETIME FROM ANIMAL_INS ORDER BY DATETIME DESC LIMIT 1 SELECT MAX(DATETIME) FROM ANIMAL_INS 최솟값 구하기 # -> Aggregate: min(animal_ins.`DATETIME`) -> Table scan on ANIMAL_INS (cost=10.25 rows=100) SELECT MIN(DATETIME) FROM ANIMAL_INS 동물 수 구하기 SELECT COUNT(1) FROM ANIMAL_INS 중복 제거하기 SELECT COUNT(DISTINCT NAME) FROM ANIMAL_INS

모든 레코드 조회하기 SELECT * FROM ANIMAL_INS ORDER BY ANIMAL_ID 역순 정렬하기 SELECT NAME, DATETIME FROM ANIMAL_INS ORDER BY ANIMAL_ID DESC # Backward index scan 아픈 동물 찾기 SELECT ANIMAL_ID, NAME FROM ANIMAL_INS WHERE INTAKE_CONDITION LIKE "Sick" ORDER BY ANIMAL_ID 어린 동물 찾기 SELECT ANIMAL_ID, NAME FROM ANIMAL_INS WHERE INTAKE_CONDITION "Aged" ORDER BY ANIMAL_ID 동물의 아이디와 이름 SELECT ANIMAL_ID, NAME FROM ANIMAL_INS ..