C# 具有附加参数的Linq Group By Into

我有以下sql结果： 表格结果

目标是按Projected和SequenceId对该结果进行分组。稍后的JSON结果应该如下所示：

[
  {
    "ProjectId": 1,
    "ProjectName": "Testprojekt 1",
    "Sequences": [
      { 
        "SequenceId": 2,
        "SequenceName": "ESN_Tauschen" 
      },
      {
        "SequenceId": 3,
        "SequenceName": "Demontage"
      }
    ]
  },
  {
    "ProjectId": 2,
    "ProjectName": "Testprojekt 2",
    "Sequences": [
      {
        "SequenceId": 3,
        "SequenceName": "Demontage"
      }
    ]
  }
]

group new { p, s } by p.ProjectId into g
let p = g.FirstOrDefault().p
select new
{
    ProjectId = g.Key,
    ProjectName = p.Name,
    Sequences =
        (from e in g
         group e.s by e.s.SequenceId into g2
         let s = g2.FirstOrDefault()
         select new
         {
             SequenceId = g2.Key,
             SequenceName = s.Name
         }).ToList()
};

我当前的linq表达式给出以下结果：

[
  {
    "ProjectId": 1,
    "Sequences": [
      2,
      3
    ]
  },
  {
    "ProjectId": 2,
    "Sequences": [
      3
    ]
  }
]

 var context = new ReworkPlace();

        var result = from p in context.Projects
                     join rs in context.ReworkStations on p.ProjectId equals rs.ProjectId
                     join l in context.ReworkStationReworkConfigurationLinkSets on rs.ReworkStationId equals
                         l.ReworkStationId
                     join rc in context.ReworkConfigurations on l.ReworkConfigurationId equals rc.ReworkConfigurationId
                     join s in context.Sequences on rc.SequenceId equals s.SequenceId
                     group s.SequenceId by p.ProjectId into g
                     select new
                     {
                         ProjectId = g.Key,
                         Sequences = g.ToList()
                     };


        return Json(result, JsonRequestBehavior.AllowGet);

我不知道如何调整linq表达式以将未分组属性如ProjectName、SequenceId和SequenceName包含到json结果中

---

如果有人能帮助我，那就太好了。

要在不完全重写查询的情况下获得所需的结果，请替换分组部分：

 group s.SequenceId by p.ProjectId into g
 select new
 {
     ProjectId = g.Key,
     Sequences = g.ToList()
 };

比如说：

[
  {
    "ProjectId": 1,
    "ProjectName": "Testprojekt 1",
    "Sequences": [
      { 
        "SequenceId": 2,
        "SequenceName": "ESN_Tauschen" 
      },
      {
        "SequenceId": 3,
        "SequenceName": "Demontage"
      }
    ]
  },
  {
    "ProjectId": 2,
    "ProjectName": "Testprojekt 2",
    "Sequences": [
      {
        "SequenceId": 3,
        "SequenceName": "Demontage"
      }
    ]
  }
]

group new { p, s } by p.ProjectId into g
let p = g.FirstOrDefault().p
select new
{
    ProjectId = g.Key,
    ProjectName = p.Name,
    Sequences =
        (from e in g
         group e.s by e.s.SequenceId into g2
         let s = g2.FirstOrDefault()
         select new
         {
             SequenceId = g2.Key,
             SequenceName = s.Name
         }).ToList()
};

诀窍是在

group

和

by

之间包含分组中所需的数据（除了

键

，该键是在

by

之后添加的）

---

若要获取其他字段，可以将它们包括在分组键中，或者如果它们与定义的分组键相同，使用

FirstOrDefault

从分组中的第一条记录中获取它们，如上例所示。

如果使用导航属性而不是加入，我认为您完全可以不使用

组。

。非常感谢伊万，您救了我一天：-）这正是您提到的工作方式！