我在C#中有多个条件的LINQ Joining语句。
var possibleSegments =
from epl in eventPotentialLegs
join sd in segmentDurations on
new {
epl.ITARequestID,
epl.ITASliceNumber,
epl.DepartAirportAfter,
epl.AirportId_Origin,
epl.AirportId_Destination
}
equals
new {
sd.ITARequestId,
sd.SliceIndex,
sd.OriginAirport,
sd.DestinationAirport
}
where
epl.DepartAirportAfter > sd.UTCDepartureTime
and
epl.ArriveAirportBy > sd.UTCArrivalTime
select new PossibleSegments{ ArrivalTime = sd.arrivalTime };
加入无效。我做错了什么?
答案 0 :(得分:95)
AFAIK你只能这样加入:
var query = from obj_i in set1
join obj_j in set2 on
new {
JoinProperty1 = obj_i.SomeField1,
JoinProperty2 = obj_i.SomeField2,
JoinProperty3 = obj_i.SomeField3,
JoinProperty4 = obj_i.SomeField4
}
equals
new {
JoinProperty1 = obj_j.SomeOtherField1,
JoinProperty2 = obj_j.SomeOtherField2,
JoinProperty3 = obj_j.SomeOtherField3,
JoinProperty4 = obj_j.SomeOtherField4
}
主要要求是: 您加入的匿名对象中的属性名称,类型和顺序必须匹配。
您不能在联接中使用ANDs OR等。只是object1等于object2。
这个LinqPad示例中更高级的内容:
class c1
{
public int someIntField;
public string someStringField;
}
class c2
{
public Int64 someInt64Property {get;set;}
private object someField;
public string someStringFunction(){return someField.ToString();}
}
void Main()
{
var set1 = new List<c1>();
var set2 = new List<c2>();
var query = from obj_i in set1
join obj_j in set2 on
new {
JoinProperty1 = (Int64) obj_i.someIntField,
JoinProperty2 = obj_i.someStringField
}
equals
new {
JoinProperty1 = obj_j.someInt64Property,
JoinProperty2 = obj_j.someStringFunction()
}
select new {obj1 = obj_i, obj2 = obj_j};
}
寻址名称和属性顺序很简单,寻址类型可以通过转换/转换/解析/调用方法等来实现。这可能并不总是适用于LINQ to EF或SQL或NHibernate,大多数方法调用肯定不起作用,在运行时会失败,所以YMMV。 这是因为它们被复制到匿名对象中的公共只读属性,因此只要表达式生成正确类型的值连接属性 - 您应该没问题。
答案 1 :(得分:5)
您的and
应该是&&
条款中的where
。
where epl.DepartAirportAfter > sd.UTCDepartureTime
and epl.ArriveAirportBy > sd.UTCArrivalTime
应该是
where epl.DepartAirportAfter > sd.UTCDepartureTime
&& epl.ArriveAirportBy > sd.UTCArrivalTime
答案 2 :(得分:0)
如果您不需要相等的对象条件,请使用交叉连接序列:
var query = from obj1 in set1
from obj2 in set2
where obj1.key1 == obj2.key2 && obj1.key3.contains(obj2.key5) [...conditions...]