cave dwellers

이것저것 개발하니 기본도 생각이 안나는구나~~ - CREATE : 생성 - ALTER : 변경 - DROP : 삭제 데이터 베이스 생성(DB명:CAVE) : CREATE DATABASE CAVE; 데이터 베이스 삭제 : DROP DATABASE IF EXIST CAVE; 현재 사용중인 데이터베이스 확인 : SELECT DATABASE(); DB선택 : USE CAVE; 타입 - CHAR(값) : 고정 값 1~255 byte 고정 길이 문자형 - VARCHAR(값) : 변동 되는 값 1~65,535byte 가변 길이 문자형 - TINYTEXT : 1~255 byte - TEXT : 1~65.535 byte - MEDIUMTEXT : 1~16,777,215 byte - LONGTEXT : 1~4,294,96..

Programmers / MySQL / Lv.3 SELECT U.USER_ID AS USER_ID, U.NICKNAME AS NICKNAME, CONCAT_WS(' ', U.CITY, U.STREET_ADDRESS1, U.STREET_ADDRESS2) AS '전체주소', CONCAT_WS('-', SUBSTRING(U.TLNO FROM 1 FOR 3), SUBSTRING(U.TLNO FROM 4 FOR 4), SUBSTRING(U.TLNO FROM 8)) AS '전화번호' FROM USED_GOODS_USER AS U INNER JOIN USED_GOODS_BOARD AS B ON U.USER_ID = B.WRITER_ID GROUP BY B.WRITER_ID HAVING COUNT(B.WRITER_ID..

Programmers / MySQL / Lv.3 SELECT O.ANIMAL_ID AS ANIMAL_ID, O.NAME AS NAME FROM ANIMAL_INS AS I RIGHT OUTER JOIN ANIMAL_OUTS AS O ON I.ANIMAL_ID = O.ANIMAL_ID WHERE I.ANIMAL_ID IS NULL ORDER BY O.ANIMAL_ID ASC; [해석] SELECT O.ANIMAL_ID AS ANIMAL_ID, O.NAME AS NAME : 입양을 간 기록은 있지만, 보호소에 들어온 기록이 없다는 것은 'ANIMAL_OUTS'의 동물 데이터를 조회 FROM ANIMAL_INS AS I RIGHT OUTER JOIN ANIMAL_OUTS AS O ON I.ANIMAL_ID =..

Programmers / MySQL / Lv.3 SELECT I.ANIMAL_ID AS ANIMAL_ID, I.NAME AS NAME FROM ANIMAL_INS AS I INNER JOIN ANIMAL_OUTS AS O ON I.ANIMAL_ID = O.ANIMAL_ID WHERE I.DATETIME > O.DATETIME ORDER BY I.DATETIME ASC; [해석] SELECT I.ANIMAL_ID AS ANIMAL_ID, I.NAME AS NAME : 동물의 아이디, 이름 출력 FROM ANIMAL_INS AS I INNER JOIN ANIMAL_OUTS AS O ON I.ANIMAL_ID = O.ANIMAL_ID : 입양 보낸 동물의 정보 필요 : 동물 정보와 입양 보낸 동물 조인 WHE..

Programmers/MySQL/Lv.3 SELECT N.CATEGORY AS CATEGORY, SUM(S.SALES) AS TOTAL_SALES FROM BOOK AS N INNER JOIN BOOK_SALES AS S ON N.BOOK_ID = S.BOOK_ID WHERE S.SALES_DATE LIKE '2022-01-%' GROUP BY N.CATEGORY ORDER BY CATEGORY ASC; [해석] SELECT N.CATEGORY AS CATEGORY, SUM(S.SALES) AS TOTAL_SALES : 카테고리 표출 : 전체 판매량이기 때문에 SUM(S.SALES)하여 판매량의 합 도출 FROM BOOK AS N INNER JOIN BOOK_SALES AS S ON N.BOOK_ID =..