Register IRestClient with different API URLs with Simple Injector

问题

I would like to register IRestClient with Simple Injector to inject it into a service:

public class ActiveCustomersService : IActiveCustomersService {
    private readonly Uri apiUri = new Uri(string.Format("{0}/{1}", ApiHelper.GetUrl(), thisApiUrl));
    private IRestClient _client;

    public ActiveCustomersService() {
        _client = new RestClient(apiUri);
    }

    public ActiveCustomersService(IRestClient client) {
        _client = client;
    }
}

However when I am trying to register it:

private void ManualServiceRegistration() {
    var container = new Container();

    container.Register<IActiveCustomersService, ActiveCustomersService>();
    container.Register<IColorPerformanceService, ColorPerformanceService>();

   // code for simple injector mvc integration 
}

I get the error

Additional information: For the container to be able to create ActiveCustomersService, it should contain exactly one public constructor, but it has 2.

So, I change the service to have a single constructor and register IRestClient, like this respectively:

public class ActiveCustomersService : IActiveCustomersService {
    private readonly Uri apiUri = new Uri(string.Format("{0}/{1}", ApiHelper.GetUrl(), thisApiUrl));
    private IRestClient _client;

    public ActiveCustomersService(IRestClient client) {
        _client = client;
    }
}

private void ManualServiceRegistration() {
    var container = new Container();

    container.RegisterSingle<IRestClient>(() => new RestClient(apiUrl))
    container.Register<IActiveCustomersService, ActiveCustomersService>();
    container.Register<IColorPerformanceService, ColorPerformanceService>();

   // code for simple injector mvc integration 
}

I realize that the apiUrl needs to be different for each of the services that I will be registering. How can I do this?

来源：https://stackoverflow.com/questions/31405336/register-irestclient-with-different-api-urls-with-simple-injector

标签

restsharp

simple-injector