我真的想做这样的事情:
Select *
from A join B on A.key = B.key join C on B.key = C.key -- propagated keys
where exists (select null from B where A.key = B.key and B.Name = "Joe") and
exists (select null from C where B.key = C.key and C.Name = "Kim")
如果使用Entity Framework4和C#,linq语句会是什么样子?
更新:
显然,.Contains()将产生"Where Exists“结果。所以,另一次尝试
(我甚至不知道这是否会编译LOL):
var inner1 = from recordB in B
where recordB.Name = "Joe"
select recordB.key;
var inner2 = from recordC in C
where recordC.Name = "Kim"
select recordC.key;
var result = from recordA in A
where inner1.Contains( recordA.key) &&
inner2.Contains( recordA.key)
select recordA;
编辑:哇,这就是实际有效的方法:
var result = from A in Products
where A.kfield1 == 1 && A.kfield2 == 2 &&
( from B in Btable
where B.otherid == "Joe" && // field I want to select by
B.kfield1 == A.kfield1 &&
B.kfield2 == A.kfield2 // Can keep adding keys here
select A.identifier // unique identity field
).Contains(A.identifier) &&
( from C in Ctable
where C.otherid == "Kim" && // field I want to select by
C.kfield1 == A.kfield1 &&
C.kfield2 == A.kfield2 // Can keep adding keys here
select A.identifier // unique identity field
).Contains(A.identifier)
select A;
这就产生了这个SQL:
SELECT [t0].[identifier], [t0].*
FROM [A] AS [t0]
WHERE ([t0].[kfield1] = @p0) AND ([t0].[kfield2] = @p1) AND (EXISTS(
SELECT NULL AS [EMPTY]
FROM [B] AS [t1]
WHERE ([t0].[identifier] = [t0].[identifier]) AND ([t1].[otherid] = @p2) AND
([t1].[kfield1] = [t0].[kfield1]) AND
([t1].[kfield2] = [t0].[kfield2]))) AND (EXISTS(
SELECT NULL AS [EMPTY]
FROM [C] AS [t2]
WHERE ([t0].[identifier] = [t0].[identifier]) AND ([t2].[otherid] = @p3) AND
([t2].[kfield1] = [t0].[kfield1]) AND
([t2].[kfiekd2] = [t0].[kfield2]) ))
这就是我想要的。注意t0.identifier = t0.identifier,它过滤掉null值,因为null不等于任何东西,包括它本身(在SQL中)
发布于 2010-11-04 00:12:48
.Any()
扩展方法通常映射到exists
。
发布于 2010-11-02 07:49:52
你有没有试过把你的exists
条件反射添加到你的joins中?
from a in context.AEntity
Join B in context.BEntity on A.Key equals B.Key && B.Name == "Joe"
Join C in context.CEntity on B.Key equals C.Key && C.Name == "Kim";
不确定这是否有效,但值得一试。
https://stackoverflow.com/questions/4074013
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