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

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。

其他回答

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);

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。

如果你在你的数据库中有一个关系设置,在BreedId上有一个外键约束,你还没有得到吗?

所以我现在可以调用:

internal Album GetAlbum(int albumId)
{
    return Albums.SingleOrDefault(a => a.AlbumID == albumId);
}

在调用它的代码中:

var album = GetAlbum(1);

foreach (Photo photo in album.Photos)
{
    [...]
}

在你的实例中,你会调用像dog。breed。breedname这样的东西-正如我说的,这依赖于你的数据库是用这些关系建立的。

正如其他人所提到的,如果存在数据库调用问题,DataLoadOptions将有助于减少数据库调用。

BreedId in the Dog table is obviously a foreign key to the corresponding row in the Breed table. If you've got your database set up properly, LINQ to SQL should automatically create an association between the two tables. The resulting Dog class will have a Breed property, and the Breed class should have a Dogs collection. Setting it up this way, you can still return IEnumerable<Dog>, which is an object that includes the breed property. The only caveat is that you need to preload the breed object along with dog objects in the query so they can be accessed after the data context has been disposed, and (as another poster has suggested) execute a method on the collection that will cause the query to be performed immediately (ToArray in this case):

public IEnumerable<Dog> GetDogs()
{
    using (var db = new DogDataContext(ConnectString))
    {
        db.LoadOptions.LoadWith<Dog>(i => i.Breed);
        return db.Dogs.ToArray();
    }

}

然后,访问每只狗的品种是微不足道的:

foreach (var dog in GetDogs())
{
    Console.WriteLine("Dog's Name: {0}", dog.Name);
    Console.WriteLine("Dog's Breed: {0}", dog.Breed.Name);        
}

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});