假设我有一个左外连接，如下所示:
from f in Foo
join b in Bar on f.Foo_Id equals b.Foo_Id into g
from result in g.DefaultIfEmpty()
select new { Foo = f, Bar = result }
如何使用扩展方法表达相同的任务?如。
Foo.GroupJoin(Bar, f => f.Foo_Id, b => b.Foo_Id, (f,b) => ???)
.Select(???)
当前回答
改进Ocelot20的答案，如果你有一个表，你只是想要0或1行，但它可以有多个，你需要对你的连接表排序:
var qry = Foos.GroupJoin(
Bars.OrderByDescending(b => b.Id),
foo => foo.Foo_Id,
bar => bar.Foo_Id,
(f, bs) => new { Foo = f, Bar = bs.FirstOrDefault() });
否则，您在join中获得的行将是随机的(或者更具体地说，无论db碰巧先找到哪一行)。
其他回答
虽然接受的答案工作，是很好的Linq对象它困扰我的SQL查询不只是一个直接的左外连接。
下面的代码依赖于允许您传递表达式并将它们调用到查询的LinqKit项目。
static IQueryable<TResult> LeftOuterJoin<TSource,TInner, TKey, TResult>(
this IQueryable<TSource> source,
IQueryable<TInner> inner,
Expression<Func<TSource,TKey>> sourceKey,
Expression<Func<TInner,TKey>> innerKey,
Expression<Func<TSource, TInner, TResult>> result
) {
return from a in source.AsExpandable()
join b in inner on sourceKey.Invoke(a) equals innerKey.Invoke(b) into c
from d in c.DefaultIfEmpty()
select result.Invoke(a,d);
}
它可以这样使用
Table1.LeftOuterJoin(Table2, x => x.Key1, x => x.Key2, (x,y) => new { x,y});
改进Ocelot20的答案，如果你有一个表，你只是想要0或1行，但它可以有多个，你需要对你的连接表排序:
var qry = Foos.GroupJoin(
Bars.OrderByDescending(b => b.Id),
foo => foo.Foo_Id,
bar => bar.Foo_Id,
(f, bs) => new { Foo = f, Bar = bs.FirstOrDefault() });
否则，您在join中获得的行将是随机的(或者更具体地说，无论db碰巧先找到哪一行)。
要实现两个数据集的连接，不需要使用组连接方法。
内连接:
var qry = Foos.SelectMany
(
foo => Bars.Where (bar => foo.Foo_id == bar.Foo_id),
(foo, bar) => new
{
Foo = foo,
Bar = bar
}
);
对于左连接，只需添加DefaultIfEmpty()
var qry = Foos.SelectMany
(
foo => Bars.Where (bar => foo.Foo_id == bar.Foo_id).DefaultIfEmpty(),
(foo, bar) => new
{
Foo = foo,
Bar = bar
}
);
EF和LINQ to SQL正确转换为SQL。 对于LINQ to Objects，最好使用GroupJoin，因为它在内部使用Lookup。但是如果您正在查询DB，那么跳过GroupJoin是AFAIK作为性能。
Personlay对我来说，这种方式比GroupJoin()更具可读性。
Marc Gravell的答案变成了一个支持IQueryable<T>接口的扩展方法，并添加了对c# 8.0 NRT的支持，如下所示:
#nullable enable
using LinqKit;
using LinqKit.Core;
using System.Linq.Expressions;
...
/// <summary>
/// Left join queryable. Linq to SQL compatible. IMPORTANT: any Includes must be put on the source collections before calling this method.
/// </summary>
public static IQueryable<TResult> LeftJoin<TOuter, TInner, TKey, TResult>(
this IQueryable<TOuter> outer,
IQueryable<TInner> inner,
Expression<Func<TOuter, TKey>> outerKeySelector,
Expression<Func<TInner, TKey>> innerKeySelector,
Expression<Func<TOuter, TInner?, TResult>> resultSelector)
{
return outer
.AsExpandable()
.GroupJoin(
inner,
outerKeySelector,
innerKeySelector,
(outerItem, innerItems) => new { outerItem, innerItems })
.SelectMany(
joinResult => joinResult.innerItems.DefaultIfEmpty(),
(joinResult, innerItem) =>
resultSelector.Invoke(joinResult.outerItem, innerItem));
}
由于这似乎是使用方法(扩展)语法的左外连接的事实上的SO问题，我想我应该在当前选择的答案之外添加一个替代方案(至少在我的经验中)，这是我更常见的答案
// Option 1: Expecting either 0 or 1 matches from the "Right"
// table (Bars in this case):
var qry = Foos.GroupJoin(
Bars,
foo => foo.Foo_Id,
bar => bar.Foo_Id,
(f,bs) => new { Foo = f, Bar = bs.SingleOrDefault() });
// Option 2: Expecting either 0 or more matches from the "Right" table
// (courtesy of currently selected answer):
var qry = Foos.GroupJoin(
Bars,
foo => foo.Foo_Id,
bar => bar.Foo_Id,
(f,bs) => new { Foo = f, Bars = bs })
.SelectMany(
fooBars => fooBars.Bars.DefaultIfEmpty(),
(x,y) => new { Foo = x.Foo, Bar = y });
要使用一个简单的数据集来显示差异(假设我们是根据值本身进行连接):
List<int> tableA = new List<int> { 1, 2, 3 };
List<int?> tableB = new List<int?> { 3, 4, 5 };
// Result using both Option 1 and 2. Option 1 would be a better choice
// if we didn't expect multiple matches in tableB.
{ A = 1, B = null }
{ A = 2, B = null }
{ A = 3, B = 3 }
List<int> tableA = new List<int> { 1, 2, 3 };
List<int?> tableB = new List<int?> { 3, 3, 4 };
// Result using Option 1 would be that an exception gets thrown on
// SingleOrDefault(), but if we use FirstOrDefault() instead to illustrate:
{ A = 1, B = null }
{ A = 2, B = null }
{ A = 3, B = 3 } // Misleading, we had multiple matches.
// Which 3 should get selected (not arbitrarily the first)?.
// Result using Option 2:
{ A = 1, B = null }
{ A = 2, B = null }
{ A = 3, B = 3 }
{ A = 3, B = 3 }
选项2对于典型的左外连接定义是正确的，但正如我前面提到的，根据数据集的不同，它通常是不必要的复杂。
