好的，所以你有一个 DbContext 和 Employees 和 Cities。每个Employee 都住在City；每个City 是零个或多个Employees 的住所。显然是使用外键的简单一对多关系。
在entity framework code-first conventions 之后，您将拥有类似于以下类的内容：
class City
{
public int Id {get; set;}
// every city Houses zero or more Employees:
public virtual ICollection<Employee> Employees {get; set;}
... // other properties
}
class Employee
{
public int Id {get; set;}
// every Employee lives in exactly one City, using foreign key
public int CityId {get; set;}
public virtual City {get; set;}
}
class MyDbContext : DbContext
{
public DbSet<City> Cities {get; set;}
public DbSet<Employee> Employees {get; set;}
}
因为我遵循实体框架代码优先约定，实体框架将能够检测表和列以及城市和员工之间的一对多关系。
只有当你想为表或列使用非默认名称时，你才需要属性或流畅的 API。
回到你的问题
给定一个Id，您需要Employee 的几个属性
Id，包括容纳此的 City 的几个属性
Employee
您可以使用连接。但是，如果您使用City 属性，那么实体框架就足够聪明，可以了解需要哪个连接。代码对读者来说看起来更自然：
var queryEmployees = myDbcontext.Employees // from the sequence of all Employees
.Where(employee => employee.Id == Id) // keep only the employee with this Id
.Select(employee => new // from the remaining employees
{ // make one new object with properties:
NameEnn = employee.R_Name_Enn,
NameArr = ut.R_Name_Arr,
AddrEnn = ut.R_Addr_Enn,
AddrArr = ut.R_Addr_Arr,
City = new // I chose to make a sub-property for city
{ // if desired, you can flatten it.
Id = employee.City.Id,
Name = employee.City.Name,
NameArr = employee.City.Name_Arr,
},
});
我预计只有一名员工使用此 ID。要获取这个 Employee 使用：
var fetchedEmployee = queryEmployees.FirstOrDefault();
或者如果你真的想要一个包含这个Employee的列表：
var fetchedEmployees = queryEmployees.ToList();
如果您真的认为联接更具可读性和可维护性 - 我对此表示怀疑 - 您可以使用内部联接获得相同的结果：
var queryEmployees = myDbcontext.Employees // from the sequence of all Employees
.Where(employee => employee.Id == Id) // keep only the employee with this Id
.Select(employee => new // join the remaining employees
.Join(myDbcontext.Cities, // with the sequence of Cities
employee => employee.CityId, // from each Employee take the CityId
city => city.Id // from each City take the Id,
(employee, city) => new // when they match
{ // make one new object with properties:
NameEnn = employee.R_Name_Enn,
NameArr = ut.R_Name_Arr,
AddrEnn = ut.R_Addr_Enn,
AddrArr = ut.R_Addr_Arr,
City = new
{
Id = city.Id,
Name = city.Name,
NameArr = city.Name_Arr,
},
});