当其中一个模型有ICollection列表时，实体框架将3个表连接起来

Question

我想加入3个模型，但其中一个有ICollection，我不知道如何管理这个。我写了简单的SQL，它的工作，但我不知道如何执行它在代码C#使用linq。我的模型。

public class Car
{
    public int IDCar { get; set; }
    public virtual CarProducent Producent { get; set; }
    public string Model { get; set; }
    public double Engine { get; set; }
    public int HorsePower { get; set; }
    public DateTime ProductionDate { get; set; }
}

public class CarProducent
{
    public int IDCarProducent { get; set; }
    public string producent { get; set; }
}
public class Klient
{
    public int IDKlient { get; set; }
    public string name{ get; set; }
    public string surname { get; set; }
    public string phone { get; set; }
    public string address { get; set; }
    public virtual ICollection<Car> Cars{ get; set; }
}

我想把这3个模型连接起来，以获得它的全部细节。我写的那种查询

var car = (  from cars in dataContext.Cars
             join prod in dataContext.CarProducent on cars.Producent.IDCarProducent equals prod.IDCarProducent
             join client in dataContext.Clients on cars.IDCar equals client.Cars. /*Here I don't know how to join */
             where cars.IDCar == id
             select new
             {
                 cars.IDCar,
                 cars.Model,
                 cars.Engine,
                 cars.HorsePower,
                 cars.ProductionDate,
                 prod.producent,
                 client.name,
                 client.surname,
                 client.address,
                 client.phone
             }).FirstOrDefault();

我只是在学习，所以任何建议都会有帮助。

Answer 1

这是简单的

from client in dataContext.Klient
from car in client.Cars
where cars.IDCar == id
select new
{
  car.IDCar,
  car.Model,
  car.Engine,
  car.HorsePower,
  car.ProductionDate,
  car.Producent.producent,
  client.name,
  client.surname,
  client.address,
  client.phone
}

如果您添加了另一个导航属性，您也可以写上

from car in dataContext.Cars
where cars.IDCar == id
select new
{
  car.IDCar,
  car.Model,
  car.Engine,
  car.HorsePower,
  car.ProductionDate,
  car.Producent.producent,
  car.Client.name,
  car.Client.surname,
  car.Client.address,
  car.Client.phone
}