无法从 Microsoft.Extensions.DependencyInjection.

Question

我正在IFoo向Microsoft.Extensions.DependencyInjection服务注册一个接口 ( )，如下所示：

servicesBuilder.AddSingleton<IFooRepository, FooInMemoryRepository>();

当我解决依赖关系时，我几乎完全不知道具体类型：

services.GetRequiredService<IFooRepository>()

在这种情况下，一切正常。然而，在极少数情况下，我需要具体的实现 ( FooInMemoryRepository)，所以我想直接解决这个问题。我试着打电话

services.GetRequiredService<FooInMemoryRepository>()

但是，这会引发异常：

System.InvalidOperationException: No service for type 'UserQuery+FooInMemoryRepository' has been registered.

   at Microsoft.Extensions.DependencyInjection.ServiceProviderServiceExtensions.GetRequiredService(IServiceProvider provider, Type serviceType)

   at Microsoft.Extensions.DependencyInjection.ServiceProviderServiceExtensions.GetRequiredService[T](IServiceProvider provider)

   at UserQuery.Main() in C:\Users\User\AppData\Local\Temp\LINQPad5\_ntmafjyh\query_eecehy.cs:line 37

示例代码（LINQPad）：

void Main()

{

    var servicesBuilder = new ServiceCollection();

    servicesBuilder.AddSingleton<IFooRepository, FooInMemoryRepository>();

    servicesBuilder.AddSingleton<IFooRepository, FooOtherRepository>();

    var services = servicesBuilder.BuildServiceProvider();

    try

    {

        var foos = services.GetServices<IFooRepository>();

        foreach(var foo in foos)

        {

            Console.WriteLine(foo.SomeMethod());

        }

        var ifoo = services.GetRequiredService<IFooRepository>();

        Console.WriteLine(ifoo.SomeMethod());

        var inMemoryRepo = services.GetRequiredService<FooInMemoryRepository>();

        Console.WriteLine(inMemoryRepo.SomeMethod());

    }

    catch(InvalidOperationException ioex)

    {

        Console.WriteLine(ioex.ToString());

    }

}

// Define other methods and classes here

public interface IFooRepository

{

    string SomeMethod();

}

public class FooInMemoryRepository : IFooRepository

{

    public string SomeMethod() => "InMemory Foo";

}

public class FooOtherRepository : IFooRepository

{

    public string SomeMethod() => "Other Foo";

}