MySQL：《学习笔记与实战》之查询实例（1）求通过率

一、求通过率

1.数据源如下

oederid	is old	createtime	state
o123	1	2018-6-1 10:25	20
o124	1	2018-6-2 1:00	20
o125	1	2018-6-1 16:00	20
o126	0	2018-6-2 6:07	20
o127	0	2018-6-1 3:00	20
o128	0	2018-6-2 0:10	10
o129	1	2018-7-1 9:00	20
o130	0	2018-7-1 0:33	10
o131	0	2018-7-1 20:00	10
o132	1	2018-7-1 10:00	20
o133	1	2018-7-1 0:11	20
o134	0	2018-7-3 2:00	20
o135	0	2018-7-3 20:00	20
o136	1	2018-7-3 19:00	10
0：新用户		10：未通过
1：老用户		20：通过

2.报表格式

报表格式
日期	新用户通过率	老用户通过率	总通过率
2018-6-1
2018-6-2

3.sql 代码

select dt,sum(new_user) as new_user_no,sum(old_user) as old_user_no,sum(user) as user_no,count(orderid) as all_user,
concat(round(sum(new_user)/count(orderid)

*100,2),'%') as new_user_rate,
concat(round(sum(old_user)/count(orderid)*100,2),'%') as old_user_rate,
concat(round(sum(user)/count(orderid)*100,2),'%') as   user_rate
from 
(select date(createtime) as dt,
    case when isold=1 and state=20   then 1 else 0 end as old_user,
    case when isold=0 and state=20  then 1  else 0 end as 

new_user,
    case when  state=20 then 1 else 0 end  as  user,orderid,state from a1 )tmp
group by dt;

 

#程老师给的代码  只提供了思路  结果是错的

select DATE_FORMAT(ctime,'%m月%d日') as 日期,sum(score-10)/count(score)*10 as 总通过率 from a group by 日期

select DATE_FORMAT(t0.ctime,'%m月%d日') as 日期,sum(case when t0.score=20 then 1 else 0 end)/count(t0.score) as 新用户通过率 from (SELECT * from a where old='0') as t0 

group by 日期; 

select DATE_FORMAT(t1.ctime,'%m月%d日') as 日期,sum(case when t1.score=20 then 1 else 0 end)/count(t1.score) as 老用户通过率 from (SELECT * from a where old='1') as t1 

group by 日期;