Question

我得到了这段代码，在这段代码中我总结了慢速和快速的驱动程序。我的问题是我必须将这笔钱与正常的司机分开。我不知道如何在这个陈述中做一个分工：

Select *
    FROM (

            Select date as Datetime, tevent.name as Event, level = case                                                                  
                           when levelname = 'High' then 'High'
                           when levelname = 'Normal'  then 'Normal'
                           when shiftname = 'Low'  then 'Low'
                           end, SUM(value) as sum

    from tCount inner join tEvent ON tCount.eventid = tevent.id
             where Name in ('Drive  Fast', 'Drive Slow')
               and date > getdate() -1
               and tevent.Name in ('E01','E02','E03','E04','E05','E06','E07','E08')
           and CalName = 'Drive'

             group by tevent.name, date, levelname
         ) as s

    PIVOT
    (
        SUM(sum)
        FOR Event IN (E01,E02,E03,E04,E05,E06,E07,E08)
    ) as p
    order by Datetime, level

然后我将相同的Select语句与普通的驱动程序放在一起：

      ...  from tCount inner join tEvent ON tCount.eventid = tevent.id
                 where Name in ('drive normal') ...

我想这样做一个部门：

(Sum('drive fast' + 'drive slow')/Sum('drive normal')) * 100

Answer 1

有一种更简单的方法可以在SQL语句中包含不同总和的不同情况：总结一个案例，如下面的百分比计算：

 Select ...
     , SUM(case Name 
            when 'drive fast' then Value 
            when 'drive slow' then value
            else 0 end) 
     / SUM(case Name 
            when 'drive normal' then value
            else 0 end) * 100 as percentage
 from ...
 where ...
 group by ...;

由于我缺少测试此代码的数据，因此我在CARS表上创建了一个查询，作为培训材料，实现了相同的原则。

select Cylinders
     , sum(case origin when 'USA' then EngineSize 
                       when 'Asia' then EngineSize 
                       else 0.0 end)
     / sum(case origin when 'Europe' then EngineSize 
                       else 0.0 end) 
     * 100 as percentage
from   sasHelp.cars
where  Cylinders in (4, 5, 6, 12)
group by Cylinders

在SQL中除以和求和

1 个答案: