左连接查询EF 6

首先，这是我使用的简化EF模型:

public partial class MaterielRoulant : IKeyed<int>
{
    public MaterielRoulant()
    {
        this.RelationsMaterielRoulantEquipement = new List<RelationMaterielRoulantEquipement>();
    }
    public int Id
    {
        get;
        set;
    }
    public virtual ICollection<RelationMaterielRoulantEquipement> RelationsMaterielRoulantEquipement
    {
        get;
        private set;
    }
}
public partial class RelationMaterielRoulantEquipement : IKeyed<int>
{
    public int EquipementId
    {
        get;
        set;
    }
    public int Id
    {
        get;
        set;
    }
    public int MaterielRoulantId
    {
        get;
        set;
    }
}
public partial class Equipement : IKeyed<int>, IEditable, IArchivable, IAuditable
{
    public int Id
    {
        get;
        set;
    }
}
public partial class Filiation : IKeyed<int>
{
    public int EquipementFilsId
    {
        get;
        set;
    }
    public int EquipementPereId
    {
        get;
        set;
    }
    public int Id
    {
        get;
        set;
    }
}

我有一个与0或1个装备相连的materielroran物品。反过来，这个设备可能有一个儿子(另一个设备)或没有。我之前使用的SQL查询是:

select * from MaterielRoulant mr
left outer join RelationMaterielRoulantEquipement rmre on mr.Id = rmre.MaterielRoulantId
left outer join Equipement e1 on rmre.EquipementId = e1.Id
left outer join Filiation f1 on e1.Id = f1.EquipementPereId
left outer join Equipement e2 on f1.EquipementFilsId = e2.Id

所以我的这个查询的EF6翻译，这不起作用，是:

var query = DataContext.Set<MaterielRoulant>()
    .Join(DataContext.Set<RelationMaterielRoulantEquipement>().DefaultIfEmpty(), x => x.Id, x => x.MaterielRoulantId, (mr, rmre) => new { mr, rmre })
    .Join(DataContext.Set<Equipement>().DefaultIfEmpty(), x => x.rmre.EquipementId, x => x.Id, (x, e1) => new { x.mr, x.rmre, e1 })
    .Join(DataContext.Set<Filiation>().DefaultIfEmpty(), x => x.e1.Id, e => e.EquipementPereId, (x, f1) => new { x.mr, x.rmre, x.e1, f1 })
    .Join(DataContext.Set<Equipement>().DefaultIfEmpty(), x => x.f1.EquipementFilsId, x => x.Id, (x, e2) => new { x.mr, x.rmre, e2 })

当我启动SQL Server Profiler时，我看到我的查询是这样翻译的:

SELECT 
    [GroupBy1].[A1] AS [C1]
    FROM ( SELECT 
        COUNT(1) AS [A1]
        FROM     [dbo].[MaterielRoulant] AS [Extent1]
        INNER JOIN  (SELECT [Extent2].[EquipementId] AS [EquipementId], [Extent2].[MaterielRoulantId] AS [MaterielRoulantId]
            FROM   ( SELECT 1 AS X ) AS [SingleRowTable1]
            LEFT OUTER JOIN [dbo].[RelationMaterielRoulantEquipement] AS [Extent2] ON 1 = 1 ) AS [Join1] ON [Extent1].[Id] = [Join1].[MaterielRoulantId]
        INNER JOIN  (SELECT [Extent3].[Id] AS [Id]
            FROM   ( SELECT 1 AS X ) AS [SingleRowTable2]
            LEFT OUTER JOIN [dbo].[Equipement] AS [Extent3] ON 1 = 1 ) AS [Join3] ON ([Join1].[EquipementId] = [Join3].[Id]) OR (([Join1].[EquipementId] IS NULL) AND ([Join3].[Id] IS NULL))
        INNER JOIN  (SELECT [Extent4].[EquipementFilsId] AS [EquipementFilsId], [Extent4].[EquipementPereId] AS [EquipementPereId]
            FROM   ( SELECT 1 AS X ) AS [SingleRowTable3]
            LEFT OUTER JOIN [dbo].[Filiation] AS [Extent4] ON 1 = 1 ) AS [Join5] ON ([Join3].[Id] = [Join5].[EquipementPereId]) OR (([Join3].[Id] IS NULL) AND ([Join5].[EquipementPereId] IS NULL))
        INNER JOIN  (SELECT [Extent5].[Id] AS [Id]
            FROM   ( SELECT 1 AS X ) AS [SingleRowTable4]
            LEFT OUTER JOIN [dbo].[Equipement] AS [Extent5] ON 1 = 1 ) AS [Join7] ON ([Join5].[EquipementFilsId] = [Join7].[Id]) OR (([Join5].[EquipementFilsId] IS NULL) AND ([Join7].[Id] IS NULL))
    )  AS [GroupBy1]

实际位置:

SELECT 
    [GroupBy1].[A1] AS [C1]
    FROM ( SELECT 
        COUNT(1) AS [A1]
        FROM     [dbo].[MaterielRoulant] AS [Extent1]
        LEFT OUTER JOIN  (SELECT [Extent2].[EquipementId] AS [EquipementId], [Extent2].[MaterielRoulantId] AS [MaterielRoulantId]
            FROM   ( SELECT 1 AS X ) AS [SingleRowTable1]
            LEFT OUTER JOIN [dbo].[RelationMaterielRoulantEquipement] AS [Extent2] ON 1 = 1 ) AS [Join1] ON [Extent1].[Id] = [Join1].[MaterielRoulantId]
        LEFT OUTER JOIN  (SELECT [Extent3].[Id] AS [Id]
            FROM   ( SELECT 1 AS X ) AS [SingleRowTable2]
            LEFT OUTER JOIN [dbo].[Equipement] AS [Extent3] ON 1 = 1 ) AS [Join3] ON ([Join1].[EquipementId] = [Join3].[Id]) OR (([Join1].[EquipementId] IS NULL) AND ([Join3].[Id] IS NULL))
        LEFT OUTER JOIN  (SELECT [Extent4].[EquipementFilsId] AS [EquipementFilsId], [Extent4].[EquipementPereId] AS [EquipementPereId]
            FROM   ( SELECT 1 AS X ) AS [SingleRowTable3]
            LEFT OUTER JOIN [dbo].[Filiation] AS [Extent4] ON 1 = 1 ) AS [Join5] ON ([Join3].[Id] = [Join5].[EquipementPereId]) OR (([Join3].[Id] IS NULL) AND ([Join5].[EquipementPereId] IS NULL))
        LEFT OUTER JOIN  (SELECT [Extent5].[Id] AS [Id]
            FROM   ( SELECT 1 AS X ) AS [SingleRowTable4]
            LEFT OUTER JOIN [dbo].[Equipement] AS [Extent5] ON 1 = 1 ) AS [Join7] ON ([Join5].[EquipementFilsId] = [Join7].[Id]) OR (([Join5].[EquipementFilsId] IS NULL) AND ([Join7].[Id] IS NULL))
    )  AS [GroupBy1]

内连接被外连接取代。

我试过使用GroupJoin，它仍然给我一个更复杂的查询错误的答案。我想保持lambda写作，因为它在我的应用程序中无处不在，但在这一点上任何工作的解决方案将是伟大的。

Thanks in advance

var q = 
    from mr in db.MaterielRoulant
    from rmre in db.RelationMaterielRoulantEquipement.Where(x => mr.Id == x.MaterielRoulantId).DefaultIfEmpty()
    from e1 in db.Equipement.Where(x => rmre.EquipementId == x.Id).DefaultIfEmpty()
    from f1 in db.Filiation.Where(x => e1.Id == x.EquipementPereId).DefaultIfEmpty()
    from e2 in db.Equipement.Where(x => f1.EquipementFilsId == x.Id).DefaultIfEmpty()
    select new {...}

db.MaterielRoulant
   .SelectMany(
      mr => 
         db.RelationMaterielRoulantEquipement
            .Where (x => mr.Id == x.MaterielRoulantId)
            .DefaultIfEmpty(), 
      (mr, rmre) => 
         new  
         {
            mr = mr, 
            rmre = rmre
         }
   )
   .SelectMany(
      temp => 
         db.Equipement
            .Where (x => rmre.EquipementId == x.Id)
            .DefaultIfEmpty(), 
      (temp, e1) =>  
        new  
        {
          mr = temp.mr, 
          rmre = temp.rmre, 
          e1 = e1
       }
   )...