如何在c# LINQ中执行左外连接到对象而不使用join-on-equal -into子句?有办法用where子句来实现吗?
正确的问题:
内连接很简单，我有一个这样的解决方案
List<JoinPair> innerFinal = (from l in lefts from r in rights where l.Key == r.Key
select new JoinPair { LeftId = l.Id, RightId = r.Id})
但是对于左外连接，我需要一个解决方案。我的是这样的，但它不工作
List< JoinPair> leftFinal = (from l in lefts from r in rights
select new JoinPair {
LeftId = l.Id,
RightId = ((l.Key==r.Key) ? r.Id : 0
})
其中JoinPair是一个类:
public class JoinPair { long leftId; long rightId; }
如果需要连接和筛选某些东西，可以在连接之外完成。可以在创建集合之后进行筛选。
在这种情况下，如果我在连接条件中这样做，我减少了返回的行。
使用三元条件(= n == null ?"__": n.MonDayNote，)
如果对象为空(因此不匹配)，则返回?后面的内容。__，在这种情况下。
否则，返回:，n.MonDayNote后面的内容。
感谢其他贡献者，这是我开始自己的问题。
var schedLocations = (from f in db.RAMS_REVENUE_LOCATIONS
join n in db.RAMS_LOCATION_PLANNED_MANNING on f.revenueCenterID equals
n.revenueCenterID into lm
from n in lm.DefaultIfEmpty()
join r in db.RAMS_LOCATION_SCHED_NOTE on f.revenueCenterID equals r.revenueCenterID
into locnotes
from r in locnotes.DefaultIfEmpty()
where f.LocID == nLocID && f.In_Use == true && f.revenueCenterID > 1000
orderby f.Areano ascending, f.Locname ascending
select new
{
Facname = f.Locname,
f.Areano,
f.revenueCenterID,
f.Locabbrev,
// MonNote = n == null ? "__" : n.MonDayNote,
MonNote = n == null ? "__" : n.MonDayNote,
TueNote = n == null ? "__" : n.TueDayNote,
WedNote = n == null ? "__" : n.WedDayNote,
ThuNote = n == null ? "__" : n.ThuDayNote,
FriNote = n == null ? "__" : n.FriDayNote,
SatNote = n == null ? "__" : n.SatDayNote,
SunNote = n == null ? "__" : n.SunDayNote,
MonEmpNbr = n == null ? 0 : n.MonEmpNbr,
TueEmpNbr = n == null ? 0 : n.TueEmpNbr,
WedEmpNbr = n == null ? 0 : n.WedEmpNbr,
ThuEmpNbr = n == null ? 0 : n.ThuEmpNbr,
FriEmpNbr = n == null ? 0 : n.FriEmpNbr,
SatEmpNbr = n == null ? 0 : n.SatEmpNbr,
SunEmpNbr = n == null ? 0 : n.SunEmpNbr,
SchedMondayDate = n == null ? dMon : n.MondaySchedDate,
LocNotes = r == null ? "Notes: N/A" : r.LocationNote
}).ToList();
Func<int, string> LambdaManning = (x) => { return x == 0 ? "" : "Manning:" + x.ToString(); };
DataTable dt_ScheduleMaster = PsuedoSchedule.Tables["ScheduleMasterWithNotes"];
var schedLocations2 = schedLocations.Where(x => x.SchedMondayDate == dMon);
在linqc#中执行左外连接
//执行左外连接
class Person
{
public string FirstName { get; set; }
public string LastName { get; set; }
}
class Child
{
public string Name { get; set; }
public Person Owner { get; set; }
}
public class JoinTest
{
public static void LeftOuterJoinExample()
{
Person magnus = new Person { FirstName = "Magnus", LastName = "Hedlund" };
Person terry = new Person { FirstName = "Terry", LastName = "Adams" };
Person charlotte = new Person { FirstName = "Charlotte", LastName = "Weiss" };
Person arlene = new Person { FirstName = "Arlene", LastName = "Huff" };
Child barley = new Child { Name = "Barley", Owner = terry };
Child boots = new Child { Name = "Boots", Owner = terry };
Child whiskers = new Child { Name = "Whiskers", Owner = charlotte };
Child bluemoon = new Child { Name = "Blue Moon", Owner = terry };
Child daisy = new Child { Name = "Daisy", Owner = magnus };
// Create two lists.
List<Person> people = new List<Person> { magnus, terry, charlotte, arlene };
List<Child> childs = new List<Child> { barley, boots, whiskers, bluemoon, daisy };
var query = from person in people
join child in childs
on person equals child.Owner into gj
from subpet in gj.DefaultIfEmpty()
select new
{
person.FirstName,
ChildName = subpet!=null? subpet.Name:"No Child"
};
// PetName = subpet?.Name ?? String.Empty };
foreach (var v in query)
{
Console.WriteLine($"{v.FirstName + ":",-25}{v.ChildName}");
}
}
// This code produces the following output:
//
// Magnus: Daisy
// Terry: Barley
// Terry: Boots
// Terry: Blue Moon
// Charlotte: Whiskers
// Arlene: No Child
https://dotnetwithhamid.blogspot.in/
现在作为一个扩展方法:
public static class LinqExt
{
public static IEnumerable<TResult> LeftOuterJoin<TLeft, TRight, TKey, TResult>(this IEnumerable<TLeft> left, IEnumerable<TRight> right, Func<TLeft, TKey> leftKey, Func<TRight, TKey> rightKey,
Func<TLeft, TRight, TResult> result)
{
return left.GroupJoin(right, leftKey, rightKey, (l, r) => new { l, r })
.SelectMany(
o => o.r.DefaultIfEmpty(),
(l, r) => new { lft= l.l, rght = r })
.Select(o => result.Invoke(o.lft, o.rght));
}
}
像平常使用join一样使用:
var contents = list.LeftOuterJoin(list2,
l => l.country,
r => r.name,
(l, r) => new { count = l.Count(), l.country, l.reason, r.people })
希望这能为您节省一些时间。
这是一般形式(已经在其他答案中提供了)
var c =
from a in alpha
join b in beta on b.field1 equals a.field1 into b_temp
from b_value in b_temp.DefaultIfEmpty()
select new { Alpha = a, Beta = b_value };
然而，这里有一个解释，我希望能澄清这实际上是什么意思!
join b in beta on b.field1 equals a.field1 into b_temp
本质上创建了一个单独的结果集b_temp，其中有效地包含了右边条目的空“rows”('b'中的条目)。
然后是下一行:
from b_value in b_temp.DefaultIfEmpty()
．．遍历该结果集，为右侧的'row'设置默认空值，并将右侧的行连接的结果设置为'b_value'的值(即，如果有匹配的记录，则为右侧的值，如果没有则为'null')。
现在，如果右边是一个单独的LINQ查询的结果，它将由匿名类型组成，它只能是'something'或'null'。如果它是一个可枚举的对象(例如，一个List -其中MyObjectB是一个有2个字段的类)，那么它可以具体说明它的属性使用什么默认的“null”值:
var c =
from a in alpha
join b in beta on b.field1 equals a.field1 into b_temp
from b_value in b_temp.DefaultIfEmpty( new MyObjectB { Field1 = String.Empty, Field2 = (DateTime?) null })
select new { Alpha = a, Beta_field1 = b_value.Field1, Beta_field2 = b_value.Field2 };
这确保了'b'本身不是空的(但它的属性可以是空的，使用您指定的默认空值)，这允许您检查b_value的属性，而不会得到b_value的空引用异常。注意，对于可空的DateTime， (DateTime?)类型，即。'nullable DateTime'必须在'DefaultIfEmpty'的规范中指定为null的'Type'(这也适用于'原生'不为空的类型，例如double, float)。
您可以通过简单地链接上述语法来执行多个左外连接。
