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

通过扩展方法的左外连接的实现如下所示

public static IEnumerable<Result> LeftJoin<TOuter, TInner, TKey, Result>(
  this IEnumerable<TOuter> outer, IEnumerable<TInner> inner
  , Func<TOuter, TKey> outerKeySelector, Func<TInner, TKey> innerKeySelector
  , Func<TOuter, TInner, Result> resultSelector, IEqualityComparer<TKey> comparer)
  {
    if (outer == null)
      throw new ArgumentException("outer");

    if (inner == null)
      throw new ArgumentException("inner");

    if (outerKeySelector == null)
      throw new ArgumentException("outerKeySelector");

    if (innerKeySelector == null)
      throw new ArgumentException("innerKeySelector");

    if (resultSelector == null)
      throw new ArgumentException("resultSelector");

    return LeftJoinImpl(outer, inner, outerKeySelector, innerKeySelector, resultSelector, comparer ?? EqualityComparer<TKey>.Default);
  }

  static IEnumerable<Result> LeftJoinImpl<TOuter, TInner, TKey, Result>(
      IEnumerable<TOuter> outer, IEnumerable<TInner> inner
      , Func<TOuter, TKey> outerKeySelector, Func<TInner, TKey> innerKeySelector
      , Func<TOuter, TInner, Result> resultSelector, IEqualityComparer<TKey> comparer)
  {
    var innerLookup = inner.ToLookup(innerKeySelector, comparer);

    foreach (var outerElment in outer)
    {
      var outerKey = outerKeySelector(outerElment);
      var innerElements = innerLookup[outerKey];

      if (innerElements.Any())
        foreach (var innerElement in innerElements)
          yield return resultSelector(outerElment, innerElement);
      else
        yield return resultSelector(outerElment, default(TInner));
     }
   }

然后,resultselector必须处理空元素。外汇。

   static void Main(string[] args)
   {
     var inner = new[] { Tuple.Create(1, "1"), Tuple.Create(2, "2"), Tuple.Create(3, "3") };
     var outer = new[] { Tuple.Create(1, "11"), Tuple.Create(2, "22") };

     var res = outer.LeftJoin(inner, item => item.Item1, item => item.Item1, (it1, it2) =>
     new { Key = it1.Item1, V1 = it1.Item2, V2 = it2 != null ? it2.Item2 : default(string) });

     foreach (var item in res)
       Console.WriteLine(string.Format("{0}, {1}, {2}", item.Key, item.V1, item.V2));
   }

其他回答

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);

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 };

通过扩展方法的左外连接的实现如下所示

public static IEnumerable<Result> LeftJoin<TOuter, TInner, TKey, Result>(
  this IEnumerable<TOuter> outer, IEnumerable<TInner> inner
  , Func<TOuter, TKey> outerKeySelector, Func<TInner, TKey> innerKeySelector
  , Func<TOuter, TInner, Result> resultSelector, IEqualityComparer<TKey> comparer)
  {
    if (outer == null)
      throw new ArgumentException("outer");

    if (inner == null)
      throw new ArgumentException("inner");

    if (outerKeySelector == null)
      throw new ArgumentException("outerKeySelector");

    if (innerKeySelector == null)
      throw new ArgumentException("innerKeySelector");

    if (resultSelector == null)
      throw new ArgumentException("resultSelector");

    return LeftJoinImpl(outer, inner, outerKeySelector, innerKeySelector, resultSelector, comparer ?? EqualityComparer<TKey>.Default);
  }

  static IEnumerable<Result> LeftJoinImpl<TOuter, TInner, TKey, Result>(
      IEnumerable<TOuter> outer, IEnumerable<TInner> inner
      , Func<TOuter, TKey> outerKeySelector, Func<TInner, TKey> innerKeySelector
      , Func<TOuter, TInner, Result> resultSelector, IEqualityComparer<TKey> comparer)
  {
    var innerLookup = inner.ToLookup(innerKeySelector, comparer);

    foreach (var outerElment in outer)
    {
      var outerKey = outerKeySelector(outerElment);
      var innerElements = innerLookup[outerKey];

      if (innerElements.Any())
        foreach (var innerElement in innerElements)
          yield return resultSelector(outerElment, innerElement);
      else
        yield return resultSelector(outerElment, default(TInner));
     }
   }

然后,resultselector必须处理空元素。外汇。

   static void Main(string[] args)
   {
     var inner = new[] { Tuple.Create(1, "1"), Tuple.Create(2, "2"), Tuple.Create(3, "3") };
     var outer = new[] { Tuple.Create(1, "11"), Tuple.Create(2, "22") };

     var res = outer.LeftJoin(inner, item => item.Item1, item => item.Item1, (it1, it2) =>
     new { Key = it1.Item1, V1 = it1.Item2, V2 = it2 != null ? it2.Item2 : default(string) });

     foreach (var item in res)
       Console.WriteLine(string.Format("{0}, {1}, {2}", item.Key, item.V1, item.V2));
   }

下面是使用方法语法的一个相当容易理解的版本:

IEnumerable<JoinPair> outerLeft =
    lefts.SelectMany(l => 
        rights.Where(r => l.Key == r.Key)
              .DefaultIfEmpty(new Item())
              .Select(r => new JoinPair { LeftId = l.Id, RightId = r.Id }));

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