Question

我有一个mysql select语句可以正常工作：

SELECT advisor AS advisor, team,        
SUM(time) -  SUM(CASE WHEN `queue` like '%Oth%' THEN time END) as case_time,
Count(distinct date) * 450 - SUM(CASE WHEN `queue` like '%Oth%' THEN time END) as time_av
FROM shift_report 
WHERE `team` = 1 and `date`>'20160701' group by `advisor` order by `advisor` ASC

我想将case_time除以time_av和* 100以获得＆＃39; actual_time＆＃39;

的百分比

我尝试了这个，但它不起作用：

round(SUM(time) -  SUM(CASE WHEN `queue` like '%Oth%' THEN time END) / 
Count(distinct date) * 450 - SUM(CASE WHEN `queue` like '%Oth%' THEN time END) * 100, 2)

任何想法??

Answer 1

我认为你错过了括号。修改后的SQL请运行并查看。

    select  round( (SUM(time) - SUM(CASE WHEN queue like '%Oth%' THEN time END))/ 
(Count(distinct date) * 450 - SUM(CASE WHEN queue like '%Oth%' THEN time END) * 
100), 2) as result FROM shift_report WHERE team = 1 and date>'20160701'

Answer 2

这有效：

round（（SUM（时间） - SUM（队列如'％Oth％'那么时间结束时））/（计数（不同日期）* 450 - SUM（队列如'％Oth％'那么时间结束））* 100,2）结果

试图找出％MYSQL，Seperate语句工作正常

2 个答案: