Generally I use some kind of conventions for this:
like
Foo.Controllers.BarController
Foo.ViewModes.BarViewModel
Foo.Repositories.BarRepository
then it's easy to register all things to the container based on the
convention. (usually you want to register ALL controllers AND all
viewModels AND all repositories so you don't even have to concern
yourself with name matching)
And if youre not always conforming to the convention youll just have few
outstading services to register explicitly.
It looks to me like you're having more of a design consistency problem
than Windsor problem
Krzysztof
Germán Schuager wrote:
> Try something like this:
>
> container.Register(
> AllTypes.Pick()
> .FromAssembly(typeof(ViewModel<>).Assembly)
> .If(c => c.Name.EndsWith("ViewModel))
> .WithService.Base() // <---- not sure about this line
> );
>
> On Sun, Nov 8, 2009 at 10:43 AM, queen3 <[email protected]
> <mailto:[email protected]>> wrote:
>
>
> Most of the examples I see for Castle Windsor to do auto-register of
> types depend on deriving from some IFoo. Or, from base class. However,
> I often have simple components (services) that just require IFoo in
> the constructor:
>
> public class AddressViewModel
> {
> public AddressViewModel(ICountryRepository countries)
> {
> this.Countries = countries.GetAll();
> }
> }
>
> Here AddressViewModel is created by ASP.NET <http://ASP.NET> MVC
> generic ActionFilter
> that transforms domain model into view model - automatically. It asks
> Windsor to create the destination view model. I have many view models,
> and I don't really want to manually register all of them in Windsor.
> From my point of view, if I ask Windsor to create AddressViewModel, it
> should check if its constructor has registered dependencies (maybe
> even nested), and if so - create them and then create AddressViewModel
> - instead of throwing "not registered".
>
> How do I make Windsor to automatically register/recognize such cases?
> I see two ways:
>
> 1. Browse all types in assembly, get their constructors, check if any
> of the constructor's parameter is registered in Windsor... and
> register it. The problem is that Service can depend on Service2 ->
> IFoo... I want CONTAINER to do this, not me.
> 2. Somehow extend Windsor (facility, etc - I'm not expert here) so
> that the above happens during "Resolve" call - so that I don't need to
> browse all the meaningless types in all assemblies.
> In both cases, however, I'll need a way to ask Windsor "Is type T your
> dependency - directly or indirectly via nested constructor?".
>
> I really hate duplication, so I don't see why I need to manually
> register something that Windsor does already know about and can
> create.
>
> Here's the code that I use now. I'm really not expert in Windsor so
> I'm not sure if the code is OK - but it works.
>
> var dependencies = container.Kernel.GetAssignableHandlers(typeof
> (object));
> // for all ViewModel classes - purely optimization, to avoid
> browsing all types
> foreach (var type in typeof(ViewModel<>).Assembly.GetTypes().Where
> (x => x.Name.EndsWith("ViewModel")))
> {
> // get all constructors and their parameters
> var ctorArgs = type.GetConstructors().SelectMany(x =>
> x.GetParameters());
> // ViewModel can be resolved if any of its ctors has Windsor-
> registered arguments
> // TODO: add checks for IList<Type>, etc... really Windsor
> should do this
> var canBeResolved = dependencies.Any(dep => ctorArgs.Any(a =>
> a.ParameterType.IsAssignableFrom
> (dep.ComponentModel.Service)));
> if (canBeResolved)
> container.AddComponent("viewModel" + type.Name, type);
> }
>
> However, I would better have Windsor (or my extension) do this during
> resolve-time, because brosing ALL types is a bit overkill in my
> opinion, even though I narrow it with name filter. Is it possible? Is
> there a better way?
>
>
>
>
> >
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