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かなり複雑です... 特定のオブジェクトに関する説明情報を含む t1 と、t1 のオブジェクトをデコードするために使用される情報を含む 2 番目のテーブル t2 の 2 つのテーブルがあります。

簡単な例を提供しようとします：

----
t1 (Name, Type, Size, Color)
----
dog1 - mammal - big  -  brown

dog2 - mammal - big  -  black

cat1 - mammal - small - black

fish1 - fish - small - gold


---
t2 (Type, Size, Color, Value)
---
mammal - "" - blue - 1

mammal - big - brown - 0.5

mammal - big - black - 0.75

mammal - small - "" - 2

fish - big - "" - 5

fish - small - "" - 0.5

fish - small - gold - 0.5

"" - "" - black - 1

"" - big - "" - 0.75

---
result(name, value)
---
dog1 - 1.25 //receives a value of 1.25 (0.5+0.75) because it hits the row of being mammal/big/brown (0.5) and the extra hit of just being big (0.75)

dog2 - 2.5 // 0.75 + 1 + 0.75 for being mammal/big/black (0.75), for being black (1), and for being big (0.75)

cat1 - 2 //2 for being mammal/small

fish1 - 1 // 0.5 + 0.5 for being fish/small and fish/small/gold

---

これは、この例に関連する私の現在のクエリですが、もちろん、いくつかの理由で機能しません。

Select t1.Name, SUM(counter.Value) as Sums
 From
 (Select *
 From t1, t2
 where t1.Type = t2.Type and t1.Size = t2.Size and t1.Color = t2.Color)
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1 に答える 1

3

これを試してください（または、空白を使用している場合は t2.XXX is null を t2.XXX = '' に置き換えます）：

declare @t1 table(name nvarchar(32), [type] nvarchar(32), size nvarchar(32), colour nvarchar(32))
declare @t2 table([type] nvarchar(32), size nvarchar(32), colour nvarchar(32), value numeric(15,9))

insert @t1
          select 'dog1','mammal','big','brown'
union all select 'dog2','mammal','big','black'
union all select 'cat1','mammal','small','black'
union all select 'fish1','fish','small','gold'

insert @t2
          select 'mammal',null,'blue',1
union all select 'mammal','big','brown',0.5
union all select 'mammal','big','black',0.75
union all select 'mammal','small',null,2
union all select 'fish','big',null,5
union all select 'fish','small',null,0.5
union all select 'fish','small','gold',0.5
union all select null,null,'black',1
union all select null,'big',null,0.75

select t1.name, sum(t2.value) Value
from @t1 t1
inner join @t2 t2
on (t1.type = t2.type or t2.type is null)
and (t1.colour = t2.colour or t2.colour is null)
and (t1.size = t2.size or t2.size is null)
group by t1.name

結果：

name    Value
cat1    3.000000000
dog1    1.250000000
dog2    2.500000000
fish1   1.000000000
于 2012-10-17T00:12:31.270 に答える