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; }

当前回答

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。

其他回答

看一下这个例子。 这个查询应该工作:

var leftFinal = from left in lefts
                join right in rights on left equals right.Left into leftRights
                from leftRight in leftRights.DefaultIfEmpty()
                select new { LeftId = left.Id, RightId = left.Key==leftRight.Key ? leftRight.Id : 0 };

这是我使用的LeftJoin实现。注意,resultSelector表达式接受2个参数:来自连接两端的一个实例。在我看到的大多数其他实现中,结果选择器只接受一个参数,这是一个具有左/右或外部/内部属性的“连接模型”。我更喜欢这个实现,因为它具有与内置Join方法相同的方法签名。它也适用于IQueryables和EF。

 var results = DbContext.Categories
      .LeftJoin(
            DbContext.Products, c => c.Id, p => p.CategoryId,
            (c, p) => new { Category = c, ProductName = p == null ? "(No Products)" : p.ProductName })
      .ToList();

public static class QueryableExtensions
{
    public static IQueryable<TResult> LeftJoin<TOuter, TInner, TKey, TResult>(
           this IQueryable<TOuter> outer, 
           IEnumerable<TInner> inner, Expression<Func<TOuter, TKey>> outerKeySelector, 
           Expression<Func<TInner, TKey>> innerKeySelector, 
           Expression<Func<TOuter, TInner, TResult>> resultSelector)
    {
        var query = outer
            .GroupJoin(inner, outerKeySelector, innerKeySelector, (o, i) => new { o, i })
            .SelectMany(o => o.i.DefaultIfEmpty(), (x, i) => new { x.o, i });
        return ApplySelector(query, x => x.o, x => x.i, resultSelector);
    }

    private static IQueryable<TResult> ApplySelector<TSource, TOuter, TInner, TResult>(
        IQueryable<TSource> source,
        Expression<Func<TSource, TOuter>> outerProperty,
        Expression<Func<TSource, TInner>> innerProperty,
        Expression<Func<TOuter, TInner, TResult>> resultSelector)
    {
        var p = Expression.Parameter(typeof(TSource), $"param_{Guid.NewGuid()}".Replace("-", string.Empty));
        Expression body = resultSelector?.Body
            .ReplaceParameter(resultSelector.Parameters[0], outerProperty.Body.ReplaceParameter(outerProperty.Parameters[0], p))
            .ReplaceParameter(resultSelector.Parameters[1], innerProperty.Body.ReplaceParameter(innerProperty.Parameters[0], p));
        var selector = Expression.Lambda<Func<TSource, TResult>>(body, p);
        return source.Select(selector);
    }
}

public static class ExpressionExtensions
{
    public static Expression ReplaceParameter(this Expression source, ParameterExpression toReplace, Expression newExpression)
        => new ReplaceParameterExpressionVisitor(toReplace, newExpression).Visit(source);
}

public class ReplaceParameterExpressionVisitor : ExpressionVisitor
{
    public ReplaceParameterExpressionVisitor(ParameterExpression toReplace, Expression replacement)
    {
        this.ToReplace = toReplace;
        this.Replacement = replacement;
    }

    public ParameterExpression ToReplace { get; }
    public Expression Replacement { get; }
    protected override Expression VisitParameter(ParameterExpression node)
        => (node == ToReplace) ? Replacement : base.VisitParameter(node);
}

看看这个例子

class Person
{
    public int ID { get; set; }
    public string FirstName { get; set; }
    public string LastName { get; set; }
    public string Phone { get; set; }
}

class Pet
{
    public string Name { get; set; }
    public Person Owner { get; set; }
}

public static void LeftOuterJoinExample()
{
    Person magnus = new Person {ID = 1, FirstName = "Magnus", LastName = "Hedlund"};
    Person terry = new Person {ID = 2, FirstName = "Terry", LastName = "Adams"};
    Person charlotte = new Person {ID = 3, FirstName = "Charlotte", LastName = "Weiss"};
    Person arlene = new Person {ID = 4, FirstName = "Arlene", LastName = "Huff"};

    Pet barley = new Pet {Name = "Barley", Owner = terry};
    Pet boots = new Pet {Name = "Boots", Owner = terry};
    Pet whiskers = new Pet {Name = "Whiskers", Owner = charlotte};
    Pet bluemoon = new Pet {Name = "Blue Moon", Owner = terry};
    Pet daisy = new Pet {Name = "Daisy", Owner = magnus};

    // Create two lists.
    List<Person> people = new List<Person> {magnus, terry, charlotte, arlene};
    List<Pet> pets = new List<Pet> {barley, boots, whiskers, bluemoon, daisy};

    var query = from person in people
        where person.ID == 4
        join pet in pets on person equals pet.Owner  into personpets
        from petOrNull in personpets.DefaultIfEmpty()
        select new { Person=person, Pet = petOrNull}; 



    foreach (var v in query )
    {
        Console.WriteLine("{0,-15}{1}", v.Person.FirstName + ":", (v.Pet == null ? "Does not Exist" : v.Pet.Name));
    }
}

// This code produces the following output:
//
// Magnus:        Daisy
// Terry:         Barley
// Terry:         Boots
// Terry:         Blue Moon
// Charlotte:     Whiskers
// Arlene:

现在你可以从左边包含元素,即使那个元素在右边没有匹配,在我们的例子中,我们检索了Arlene,即使他在右边没有匹配

这是参考资料

如何:执行左外连接(c#编程指南)

我想补充的是,如果您获得MoreLinq扩展,现在支持同质和异构左连接

http://morelinq.github.io/2.8/ref/api/html/Overload_MoreLinq_MoreEnumerable_LeftJoin.htm

例子:

//Pretend a ClientCompany object and an Employee object both have a ClientCompanyID key on them

return DataContext.ClientCompany
    .LeftJoin(DataContext.Employees,                         //Table being joined
        company => company.ClientCompanyID,                  //First key
        employee => employee.ClientCompanyID,                //Second Key
        company => new {company, employee = (Employee)null}, //Result selector when there isn't a match
        (company, employee) => new { company, employee });   //Result selector when there is a match

编辑:

回想起来,这可能会起作用,但它将IQueryable转换为IEnumerable,因为morelinq不会将查询转换为SQL。

您可以使用如下所述的GroupJoin: https://stackoverflow.com/a/24273804/4251433

这将确保它仍然是一个IQueryable,以防您以后需要对它进行进一步的逻辑操作。

下面是一个版本的扩展方法解决方案,使用IQueryable代替IEnumerable

public class OuterJoinResult<TLeft, TRight>
{
    public TLeft LeftValue { get; set; }
    public TRight RightValue { get; set; }
}

public static IQueryable<TResult> LeftOuterJoin<TLeft, TRight, TKey, TResult>(this IQueryable<TLeft> left, IQueryable<TRight> right, Expression<Func<TLeft, TKey>> leftKey, Expression<Func<TRight, TKey>> rightKey, Expression<Func<OuterJoinResult<TLeft, TRight>, TResult>> result)
{
    return left.GroupJoin(right, leftKey, rightKey, (l, r) => new { l, r })
          .SelectMany(o => o.r.DefaultIfEmpty(), (l, r) => new OuterJoinResult<TLeft, TRight> { LeftValue = l.l, RightValue = r })
          .Select(result);
}