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

下面是一个例子,如果你需要连接2个以上的表:

from d in context.dc_tpatient_bookingd
join bookingm in context.dc_tpatient_bookingm 
     on d.bookingid equals bookingm.bookingid into bookingmGroup
from m in bookingmGroup.DefaultIfEmpty()
join patient in dc_tpatient
     on m.prid equals patient.prid into patientGroup
from p in patientGroup.DefaultIfEmpty()

裁判:https://stackoverflow.com/a/17142392/2343

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)

class Program
{
    List<Employee> listOfEmp = new List<Employee>();
    List<Department> listOfDepart = new List<Department>();

    public Program()
    {
        listOfDepart = new List<Department>(){
            new Department { Id = 1, DeptName = "DEV" },
            new Department { Id = 2, DeptName = "QA" },
            new Department { Id = 3, DeptName = "BUILD" },
            new Department { Id = 4, DeptName = "SIT" }
        };


        listOfEmp = new List<Employee>(){
            new Employee { Empid = 1, Name = "Manikandan",DepartmentId=1 },
            new Employee { Empid = 2, Name = "Manoj" ,DepartmentId=1},
            new Employee { Empid = 3, Name = "Yokesh" ,DepartmentId=0},
            new Employee { Empid = 3, Name = "Purusotham",DepartmentId=0}
        };

    }
    static void Main(string[] args)
    {
        Program ob = new Program();
        ob.LeftJoin();
        Console.ReadLine();
    }

    private void LeftJoin()
    {
        listOfEmp.GroupJoin(listOfDepart.DefaultIfEmpty(), x => x.DepartmentId, y => y.Id, (x, y) => new { EmpId = x.Empid, EmpName = x.Name, Dpt = y.FirstOrDefault() != null ? y.FirstOrDefault().DeptName : null }).ToList().ForEach
            (z =>
            {
                Console.WriteLine("Empid:{0} EmpName:{1} Dept:{2}", z.EmpId, z.EmpName, z.Dpt);
            });
    }
}

class Employee
{
    public int Empid { get; set; }
    public string Name { get; set; }
    public int DepartmentId { get; set; }
}

class Department
{
    public int Id { get; set; }
    public string DeptName { get; set; }
}

输出

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);

扩展方法,类似于使用join语法的左连接

public static class LinQExtensions
{
    public static IEnumerable<TResult> LeftJoin<TOuter, TInner, TKey, TResult>(
        this IEnumerable<TOuter> outer, IEnumerable<TInner> inner, 
        Func<TOuter, TKey> outerKeySelector, 
        Func<TInner, TKey> innerKeySelector, 
        Func<TOuter, TInner, TResult> resultSelector)
    {
        return outer.GroupJoin(
            inner, 
            outerKeySelector, 
            innerKeySelector,
            (outerElement, innerElements) => resultSelector(outerElement, innerElements.FirstOrDefault()));
    }
}

我刚刚在。net内核中编写了它,它看起来像预期的那样工作。

小测试:

        var Ids = new List<int> { 1, 2, 3, 4};
        var items = new List<Tuple<int, string>>
        {
            new Tuple<int, string>(1,"a"),
            new Tuple<int, string>(2,"b"),
            new Tuple<int, string>(4,"d"),
            new Tuple<int, string>(5,"e"),
        };

        var result = Ids.LeftJoin(
            items,
            id => id,
            item => item.Item1,
            (id, item) => item ?? new Tuple<int, string>(id, "not found"));

        result.ToList()
        Count = 4
        [0]: {(1, a)}
        [1]: {(2, b)}
        [2]: {(3, not found)}
        [3]: {(4, d)}