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

如果使用数据库驱动的LINQ提供程序,则可以这样编写可读性明显更好的左外连接:

from c in categories 
from p in products.Where(c == p.Category).DefaultIfEmpty()

如果省略DefaultIfEmpty(),则有一个内部连接。

拿一个公认的答案来说:

  from c in categories
    join p in products on c equals p.Category into ps
    from p in ps.DefaultIfEmpty()

这个语法非常混乱,当你想要左连接MULTIPLE表时,它是如何工作的并不清楚。

请注意 应该注意的是,Repo.whatever.Where(condition). defaultifempty()中的from alias与外部应用/左连接-lateral相同,任何(像样的)数据库优化器都完全能够将其转换为左连接,只要您不引入每行值(也就是实际的外部应用)。不要在Linq-2-Objects中这样做(因为当你使用Linq-to-Objects时没有DB-optimizer)。

详细的例子

var query2 = (
    from users in Repo.T_User
    from mappings in Repo.T_User_Group
         .Where(mapping => mapping.USRGRP_USR == users.USR_ID)
         .DefaultIfEmpty() // <== makes join left join
    from groups in Repo.T_Group
         .Where(gruppe => gruppe.GRP_ID == mappings.USRGRP_GRP)
         .DefaultIfEmpty() // <== makes join left join

    // where users.USR_Name.Contains(keyword)
    // || mappings.USRGRP_USR.Equals(666)  
    // || mappings.USRGRP_USR == 666 
    // || groups.Name.Contains(keyword)

    select new
    {
         UserId = users.USR_ID
        ,UserName = users.USR_User
        ,UserGroupId = groups.ID
        ,GroupName = groups.Name
    }

);


var xy = (query2).ToList();

当与LINQ 2 SQL一起使用时,它将很好地翻译为以下非常清晰的SQL查询:

SELECT 
     users.USR_ID AS UserId 
    ,users.USR_User AS UserName 
    ,groups.ID AS UserGroupId 
    ,groups.Name AS GroupName 
FROM T_User AS users

LEFT JOIN T_User_Group AS mappings
   ON mappings.USRGRP_USR = users.USR_ID

LEFT JOIN T_Group AS groups
    ON groups.GRP_ID == mappings.USRGRP_GRP

编辑:

参见" 将SQL Server查询转换为Linq查询 对于一个更复杂的例子。

此外,如果你在LINQ -2- objects(而不是LINQ -2-SQL)中这样做,你应该用老式的方式来做(因为LINQ to SQL正确地将此转换为连接操作,但在对象上,这种方法强制完全扫描,并且不利用索引搜索,无论如何…):

    var query2 = (
    from users in Repo.T_Benutzer
    join mappings in Repo.T_Benutzer_Benutzergruppen on mappings.BEBG_BE equals users.BE_ID into tmpMapp
    join groups in Repo.T_Benutzergruppen on groups.ID equals mappings.BEBG_BG into tmpGroups
    from mappings in tmpMapp.DefaultIfEmpty()
    from groups in tmpGroups.DefaultIfEmpty()
    select new
    {
         UserId = users.BE_ID
        ,UserName = users.BE_User
        ,UserGroupId = mappings.BEBG_BG
        ,GroupName = groups.Name
    }

);

其他回答

如“Perform left outer joins”中所述:

var q =
    from c in categories
    join pt in products on c.Category equals pt.Category into ps_jointable
    from p in ps_jointable.DefaultIfEmpty()
    select new { Category = c, ProductName = p == null ? "(No products)" : p.ProductName };

使用lambda表达式

db.Categories    
  .GroupJoin(db.Products,
      Category => Category.CategoryId,
      Product => Product.CategoryId,
      (x, y) => new { Category = x, Products = y })
  .SelectMany(
      xy => xy.Products.DefaultIfEmpty(),
      (x, y) => new { Category = x.Category, Product = y })
  .Select(s => new
  {
      CategoryName = s.Category.Name,     
      ProductName = s.Product.Name   
  });

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);
}
(from a in db.Assignments
     join b in db.Deliveryboys on a.AssignTo equals b.EmployeeId  

     //from d in eGroup.DefaultIfEmpty()
     join  c in  db.Deliveryboys on a.DeliverTo equals c.EmployeeId into eGroup2
     from e in eGroup2.DefaultIfEmpty()
     where (a.Collected == false)
     select new
     {
         OrderId = a.OrderId,
         DeliveryBoyID = a.AssignTo,
         AssignedBoyName = b.Name,
         Assigndate = a.Assigndate,
         Collected = a.Collected,
         CollectedDate = a.CollectedDate,
         CollectionBagNo = a.CollectionBagNo,
         DeliverTo = e == null ? "Null" : e.Name,
         DeliverDate = a.DeliverDate,
         DeliverBagNo = a.DeliverBagNo,
         Delivered = a.Delivered

     });

根据我对类似问题的回答,在这里:

Linq到SQL左外连接使用Lambda语法和连接2列(复合连接键)

获得代码在这里,或克隆我的github回购,并发挥!

查询:

        var petOwners =
            from person in People
            join pet in Pets
            on new
            {
                person.Id,
                person.Age,
            }
            equals new
            {
                pet.Id,
                Age = pet.Age * 2, // owner is twice age of pet
            }
            into pets
            from pet in pets.DefaultIfEmpty()
            select new PetOwner
            {
                Person = person,
                Pet = pet,
            };

λ:

        var petOwners = People.GroupJoin(
            Pets,
            person => new { person.Id, person.Age },
            pet => new { pet.Id, Age = pet.Age * 2 },
            (person, pet) => new
            {
                Person = person,
                Pets = pet,
            }).SelectMany(
            pet => pet.Pets.DefaultIfEmpty(),
            (people, pet) => new
            {
                people.Person,
                Pet = pet,
            });