Question

在以下查询中，我获得了＃1111的Group函数的无效使用-坦率地说，我一次又一次地检查它，没有发现任何不正确的地方。有什么我想念的吗？

让Sublime中的查询检查淫秽括号的数量是否正确并且也结婚了...

SELECT CURRENT_TIME,
s.`kickoff`,
s.`home`,s.`away`,
Sum(((i.010H+i.010A+i.1020H+i.1020A+i.2030H+i.2030A+i.3040H+i.3040A+i.4045H+i.4045A)/(count(i.home)))*100) as averageFHGHome,
sum(((case when i.010H+i.010A+i.1020H+i.1020A+i.2030H+i.2030A+i.3040H+i.3040A+i.4045H+i.4045A>0.5 then 1 else 0 end)/(count(i.home)))*100) as occurFHGHome,
sum(((case when i.010H+i.010A+i.1020H+i.1020A+i.2030H+i.2030A+i.3040H+i.3040A+i.4045H+i.4045A>1.5 then 1 else 0 end)/(count(i.home)))*100) as occur2FHGHome,
Sum(((i.4560H+i.6070H+i.7080H+i.80FTH+i.4560A+i.6070A+i.7080A+i.80FTA)/(count(i.home)))*100) as averageSHGHome,
sum(((case when i.4560H+i.6070H+i.7080H+i.80FTH+i.4560A+i.6070A+i.7080A+i.80FTA>0.5 then 1 else 0 end)/(count(i.home)))*100) as occurSHGHome,
sum(((case when i.4560H+i.6070H+i.7080H+i.80FTH+i.4560A+i.6070A+i.7080A+i.80FTA>1.5 then 1 else 0 end)/(count(i.home)))*100) as occur2SHGHome,
Sum(((i.010H+i.010A+i.1020H+i.1020A+i.2030H+i.2030A+i.3040H+i.3040A+i.4045H+i.4045A+i.4560H+i.6070H+i.7080H+i.80FTH+i.4560A+i.6070A+i.7080A+i.80FTA)/(count(i.home)))*100) as averageMG,
sum(((case when i.010H+i.010A+i.1020H+i.1020A+i.2030H+i.2030A+i.3040H+i.3040A+i.4045H+i.4045A+i.4560H+i.6070H+i.7080H+i.80FTH+i.4560A+i.6070A+i.7080A+i.80FTA>0.5 then 1 else 0 end)/(count(i.home)))*100) as occur1MG,
sum(((case when i.010H+i.010A+i.1020H+i.1020A+i.2030H+i.2030A+i.3040H+i.3040A+i.4045H+i.4045A+i.4560H+i.6070H+i.7080H+i.80FTH+i.4560A+i.6070A+i.7080A+i.80FTA>1.5 then 1 else 0 end)/(count(i.home)))*100) as occur2MG,
sum(((case when i.010H+i.010A+i.1020H+i.1020A+i.2030H+i.2030A+i.3040H+i.3040A+i.4045H+i.4045A+i.4560H+i.6070H+i.7080H+i.80FTH+i.4560A+i.6070A+i.7080A+i.80FTA>2.5 then 1 else 0 end)/(count(i.home)))*100) as occur3MG
from saturday s
inner join ITACornerNew i on s.home=i.Home
group by s.home
HAVING s.kickoff > subTIME(CURRENT_TIME, '01:45:00') and s.kickoff < CURRENT_TIME
order by s.kickoff ASC

Answer 1

由于您的“淫秽括号” ，您最终获得了无效

Sum(...count(...)...)

这是聚合的聚合，而您可能正在寻找类似

的东西

Sum(...)/count(...)

与您当前的问题没有直接关系，但另外，如果s.home不是您的主键，则可能会得到不正确的结果或an error message，因为s.kickoff和{{1} }不属于您的s.away。而且，正如@Strawberry已经提到的那样，您应该强烈考虑重新设计group by。

＃1111-组功能的无效使用...是吗？

1 个答案: