我傳遞的兩個日期,需要GROUP BY的結果通過這兩個日期,但不能這樣做,因爲我得到的錯誤如何通過外部引用或其他方式進行分組?
消息164,級別15,狀態1,行24每個GROUP BY表達式必須 包含至少一個不是外部引用的列。
查詢如下:
declare @sd datetime ='2012-07-01 00:00:00.000' ,
@ed datetime ='2012-09-30 00:00:00.000' ;
select @sd,@ed,
count(i.id)as count,
sum(case when oi.rating <50 then 1 else 0 end) as unfav,
sum(case when oi.Rating =50 then 1 else 0 end) as neu,
sum(case when oi.Rating >50 then 1 else 0 end) as fav,
avg(oi.Rating)as 'Av Rating'
from Items i (nolock)
inner join ItemOrganisations oi (nolock) on i.ID= oi.ItemID
inner join Lookup_ItemTypes it (nolock) on it.ID = i.ItemTypeID
inner join Batches b (nolock) on b.ID=i.BatchID
inner join Lookup_ItemStatus lis (nolock) on lis.ID = i.StatusID
inner join Lookup_BatchStatus lbs (nolock) on lbs.ID = b.StatusID
inner join Lookup_BatchTypes bt on bt.id = b.Typeid
where lbs.Name = 'Completed by Analyst' or lbs.Name='Delivered/Imported into Neptune Online'
and lis.Name = 'Complete'
and i.IsRelevant = 1
and bt.Name = 'Live'
group by @sd,@ed
having i.OverrideDate between @sd and @ed
如果我這個不羣是結果我得到這是錯誤的:
2011-01-01 00:00:00.000 2011-01-31 00:00:00.000 1 0 0 1 55
2011-01-01 00:00:00.000 2011-01-31 00:00:00.000 7 1 0 1 50
2011-01-01 00:00:00.000 2011-01-31 00:00:00.000 7 1 0 0 20
2011-01-01 00:00:00.000 2011-01-31 00:00:00.000 1 0 0 0 NULL
2011-01-01 00:00:00.000 2011-01-31 00:00:00.000 8 1 0 6 66
2011-01-01 00:00:00.000 2011-01-31 00:00:00.000 1 1 0 0 10
2011-02-01 00:00:00.000 2011-02-28 00:00:00.000 1 0 0 1 55
2011-02-01 00:00:00.000 2011-02-28 00:00:00.000 7 1 0 1 50
2011-02-01 00:00:00.000 2011-02-28 00:00:00.000 7 1 0 0 20
2011-02-01 00:00:00.000 2011-02-28 00:00:00.000 1 0 0 0 NULL
2011-02-01 00:00:00.000 2011-02-28 00:00:00.000 8 1 0 6 66
2011-02-01 00:00:00.000 2011-02-28 00:00:00.000 1 1 0 0 10
2011-03-01 00:00:00.000 2011-03-31 00:00:00.000 1 0 0 1 55
2011-03-01 00:00:00.000 2011-03-31 00:00:00.000 7 1 0 1 50
2011-03-01 00:00:00.000 2011-03-31 00:00:00.000 7 1 0 0 20
2011-03-01 00:00:00.000 2011-03-31 00:00:00.000 1 0 0 0 NULL
2011-03-01 00:00:00.000 2011-03-31 00:00:00.000 8 1 0 6 66
2011-03-01 00:00:00.000 2011-03-31 00:00:00.000 1 1 0 0 10
2011-04-01 00:00:00.000 2011-04-30 00:00:00.000 1 0 0 1 55
2011-04-01 00:00:00.000 2011-04-30 00:00:00.000 7 1 0 1 50
2011-04-01 00:00:00.000 2011-04-30 00:00:00.000 7 1 0 0 20
沒有理由這樣做,請解釋一下你的結果期望與您進行分組的結果邏輯 – jazzytomato 2013-02-26 16:48:59
請參閱edit.thanks – Xerxes 2013-02-26 16:53:46
確定這些結果來自同一個查詢?你如何在前兩欄的「@ sd,@ ed」中得到不同的結果? – Kaf 2013-02-26 17:03:53