Question

我写了一个查询，增加了上周的工作时间。

select laborcode, sum(regularhrs) as TOTALACTUALS
from labtrans
where (laborcode='a' OR laborcode='b' OR laborcode='c' OR laborcode='d'
OR laborcode='e' OR laborcode='f') and (startdate BETWEEN DATEADD(day, DATEDIFF(day, 7, getdate()), 0) and DATEADD(day, DATEDIFF(day, 0, getdate()), 0))
group by laborcode;

让我们说劳动＆＃34; a＆＃34;距上周0小时，我的结果是：

b   25,5
c   37,25
d   24
e   48,5
f   25,5

但我想得到劳动＆＃34; a＆＃34;但也有空值。例如：

a   0 (or null)
b   25,5
c   37,25
d   24
e   48,5
f   25,5

Answer 1

将按值分组的左连接到labourcodes列表：

select distinct b.labourcode, coalesce(a.TotalHours,0) as TotalHours
from labtrans b
left join 
(
    select a.laborcode, sum(a.regularhrs) as 
    from labtrans a
    where a.labourcode in ('a','b','c','d','e','f') 
    and a.startdate BETWEEN DATEADD(day, DATEDIFF(day, 7, getdate()), 0) and DATEADD(day, DATEDIFF(day, 0, getdate()), 0)
    group by a.laborcode
)
on a.labourcode = b.labourcode

Answer 2

使用条件SUM

select laborcode
  , sum(CASE WHEN startdate BETWEEN DATEADD(day, DATEDIFF(day, 7, getdate()), 0) and DATEADD(day, DATEDIFF(day, 0, getdate()), 0) THEN regularhrs END) as TOTALACTUALS
from labtrans
where laborcode in('a','b','c','d','e','f') 
group by laborcode;

或者加入一个预定列表

SELECT l.laborcode, sum(d.regularhrs) as TOTALACTUALS
from (
     values ('a'),('b'),('c'),('d'),('e'),('f')  
     ) l(laborcode)
left join labtrans d on d.laborcode = l.laborcode and (d.startdate BETWEEN DATEADD(day, DATEDIFF(day, 7, getdate()), 0) and DATEADD(day, DATEDIFF(day, 0, getdate()), 0))
group by l.laborcode

如何返回sum函数的null值？

2 个答案: