목록데이터베이스/MYSQL (1)
하루종일 삽질
MYSQL 그룹별 연속된 날짜 갯수 세기
SET @k=0; SELECT member_no, biz_date, COUNT(group_date) FROM ( SELECT member_no, biz_date, @k:=@k+1 AS row_num, ADDDATE(biz_date, -@k) AS group_date FROM ( SELECT a.member_no, a.biz_date, SUM(b.pay_amount) FROM bizshowings AS a RIGHT JOIN bizshowings_pay AS b ON a.no = b.biz_no WHERE a.biz_date > '2019-12-12' GROUP BY a.member_no, DATE(a.biz_date) ORDER BY a.member_no ) AS aa GROUP BY member_no,..
데이터베이스/MYSQL
2020. 1. 15. 16:30