使用下面的简单示例,使用Linq to SQL从多个表返回结果的最佳方法是什么?

假设我有两个表:

Dogs:   Name, Age, BreedId
Breeds: BreedId, BreedName

我想返回所有的狗与他们的育种名称。我应该让所有的狗使用这样的东西,没有问题:

public IQueryable<Dog> GetDogs()
{
    var db = new DogDataContext(ConnectString);
    var result = from d in db.Dogs
                 join b in db.Breeds on d.BreedId equals b.BreedId
                 select d;
    return result;
}

但如果我想要有品种的狗,并尝试这样做,我有问题:

public IQueryable<Dog> GetDogsWithBreedNames()
{
    var db = new DogDataContext(ConnectString);
    var result = from d in db.Dogs
                 join b in db.Breeds on d.BreedId equals b.BreedId
                 select new
                        {
                            Name = d.Name,
                            BreedName = b.BreedName
                        };
    return result;
}

现在我意识到编译器不让我返回一组匿名类型,因为它期待狗,但有没有一种方法来返回这个而不必创建一个自定义类型?或者我必须为DogsWithBreedNames创建自己的类,并在选择中指定该类型?或者还有其他更简单的方法吗?


当前回答

如果主要的想法是让SQL选择语句发送到数据库服务器只有必需的字段,而不是所有的实体字段,那么你可以这样做:

public class Class1
{
    public IList<Car> getCarsByProjectionOnSmallNumberOfProperties()
    {

        try
        {
            //Get the SQL Context:
            CompanyPossessionsDAL.POCOContext.CompanyPossessionsContext dbContext 
                = new CompanyPossessionsDAL.POCOContext.CompanyPossessionsContext();

            //Specify the Context of your main entity e.g. Car:
            var oDBQuery = dbContext.Set<Car>();

            //Project on some of its fields, so the created select statment that is
            // sent to the database server, will have only the required fields By making a new anonymouse type
            var queryProjectedOnSmallSetOfProperties 
                = from x in oDBQuery
                    select new
                    {
                        x.carNo,
                        x.eName,
                        x.aName
                    };

            //Convert the anonymouse type back to the main entity e.g. Car
            var queryConvertAnonymousToOriginal 
                = from x in queryProjectedOnSmallSetOfProperties
                    select new Car
                    {
                        carNo = x.carNo,
                        eName = x.eName,
                        aName = x.aName
                    };

            //return the IList<Car> that is wanted
            var lst = queryConvertAnonymousToOriginal.ToList();
            return lst;

        }
        catch (Exception ex)
        {
            System.Diagnostics.Debug.WriteLine(ex.ToString());
            throw;
        }
    }
}

其他回答

如果你要归还《Dogs》,你应该这样做:

public IQueryable<Dog> GetDogsWithBreedNames()
{
    var db = new DogDataContext(ConnectString);
    return from d in db.Dogs
           join b in db.Breeds on d.BreedId equals b.BreedId
           select d;
}

如果您希望Breed立即加载而不是延迟加载,只需使用适当的DataLoadOptions构造。

必须首先使用ToList()方法从数据库中获取行,然后选择项作为类。 试试这个:

public partial class Dog {
    public string BreedName  { get; set; }}

List<Dog> GetDogsWithBreedNames(){
    var db = new DogDataContext(ConnectString);
    var result = (from d in db.Dogs
                  join b in db.Breeds on d.BreedId equals b.BreedId
                  select new
                  {
                      Name = d.Name,
                      BreedName = b.BreedName
                  }).ToList()
                    .Select(x=> 
                          new Dog{
                              Name = x.Name,
                              BreedName = x.BreedName,
                          }).ToList();
return result;}

因此,诀窍首先是ToList()。它是立即进行查询并从数据库中获取数据。第二个技巧是选择项并使用对象初始化器生成加载项的新对象。

希望这能有所帮助。

尝试这样获取动态数据。您可以转换List<>的代码

public object GetDogsWithBreedNames()
{
    var db = new DogDataContext(ConnectString);
    var result = from d in db.Dogs
                 join b in db.Breeds on d.BreedId equals b.BreedId
                 select new
                        {
                            Name = d.Name,
                            BreedName = b.BreedName
                        };
    return result.FirstOrDefault();
}

dynamic dogInfo=GetDogsWithBreedNames();
var name = dogInfo.GetType().GetProperty("Name").GetValue(dogInfo, null);
var breedName = dogInfo.GetType().GetProperty("BreedName").GetValue(dogInfo, null);

我倾向于这样的模式:

public class DogWithBreed
{
    public Dog Dog { get; set; }
    public string BreedName  { get; set; }
}

public IQueryable<DogWithBreed> GetDogsWithBreedNames()
{
    var db = new DogDataContext(ConnectString);
    var result = from d in db.Dogs
                 join b in db.Breeds on d.BreedId equals b.BreedId
                 select new DogWithBreed()
                        {
                            Dog = d,
                            BreedName = b.BreedName
                        };
    return result;
}

这意味着你有一个额外的类,但它是快速和容易编码,易于扩展,可重用和类型安全。

这并没有完全回答你的问题,但谷歌根据关键字引导我到这里。这是从列表中查询匿名类型的方法:

var anon = model.MyType.Select(x => new { x.Item1, x.Item2});