Question

我有一张类似下面的表......

我需要添加具有完全相同日期的每一行的值。（例如前两行将等于'5'，接下来的两行，'5'，接下来的两行，'8'）

StoretCode   LocName    ParName   DateTime       Value
95010-A-9    AB 901_4   On Line   201305160000   1
NULL         AB 905_8   On Line   201305160000   4
95010-A-9    AB 901_4   On Line   201305170000   2
NULL         AB 905_8   On Line   201305170000   3
95010-A-9    AB 901_4   On Line   201305180000   4
NULL         AB 905_8   On Line   201305180000   4

结果还应该有一个新的LocName ...所以它应该看起来像......

StoretCode   LocName  ParName  DateTime      Value
95010-A-9    AERBAS   On Line  201305160000  5
95010-A-9    AERBAS   On Line  201305170000  5
95010-A-9    AERBAS   On Line  201305180000  8

我试过......

Select 'AB 901_4','AB 905_8', Sum(Value)
from dbo.DATA

但结果如预期的那样......一行总计所有行......

也试过.....

Select 'AB 901_4','AB 905_8', Sum(Value)
from dbo.DATA
Where LocName, DateTime = LocName, DateTime

但结果是......

Msg 4145，Level 15，State 1，Line 3
在预期条件的上下文中指定的非布尔类型的表达式，靠近'，'。

尝试了其他几个但结果类似于上面的非布尔消息...

Answer 1

Select StoretCode , 'AERBAS' as LocName , ParName , DATA.DateTime, datecount.count as [value]
from dbo.DATA 
join ( select DateTime, count(*) as count from dbo.DATA group by DateTime ) datecount 
  on data.DateTime = datecount.DateTime 
where date.StoretCode is not null

如果你想要LocName的硬编码值，那么'LocName'

Answer 2

也许是这些方面的东西？

Select 
  max(StoretCode), 
  'AERBAS' as Locname, 
  max(ParName), 
  DateTime, 
  Sum(Value)
from dbo.DATA
group by DateTime
order by DateTime

T-SQL select语句... SUM和HAVING问题

2 个答案: