MySQL数据库查询问题,高分求解.

显示全部楼层 · 2012-11-13 11:20:46

知识点：group by 汇总函数sum/count函数难点：多条件汇总答案：%百分比，不要在sql中求，建议你用php取到结果，再算。SELECT class,SUM(id), FROM 你的表名GROUP BY class,sign_date其他：你的json不是这样用的。要用php得到结果后，再用php_json来输出json格式。...

千问 · 2012-11-13 11:20:46

你这个签到率算不出来啊，看不到某个班级的学生总人数。签到数倒是可以算出来滴SELECTCLASS,SIGN_DATE,COUNT(*)FROMTABNAMEGROUP BYCLASS,SIGN_DATE酱紫就可以看到某个班在某天的签到总人数鸟，但是班级人员的总人数你需要再想办法统计下，从签到...

千问 · 2012-11-13 11:20:46

MSSQL的，自己转一下SELECT T1.class, T2.NCOUNT, T3.SCOUNT/T2.NCOUNT AS 签到率 FROM (SELECT DISCINT class FROM T)T1 --班级ViewLEFT JOIN (SELECT COUNT(T.name) AS NCOUNT, T.class
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千问 · 2012-11-13 11:20:46

select class,count(sign_date) as 签到数,count(sign_date)/count(name) as 签到率 from table group by class;...

千问 · 2012-11-13 11:20:46

select a.class,(b.qdrs) as '签到人数',,b.qdrs/count(a.name) as '签到率'from table ainner join(select class,count(sign_date) as 'qdrs' from table where sign_date not null group by cl...