I am registering many repositories which are located in the same assembly by using the following code:
IoCContainer.Register(AllTypes.FromAssemblyContaining<RepositoryOne>).BasedOn(typeof(IRepository<>)).WithService.AllInterfaces().Configure(c => c.LifeStyle.Singleton));
Because I want to have influence on the creation of these repositories, I am trying to use UsingFactoryMethod() with it. Using this method is simple when I register every repository separately like
IoCContainer.Register(Component.For<IRepositoryOne>().ImplementedBy<RepositoryOne>().LifeStyle.Singleton.UsingFactoryMethod(() => Factory.New<RepositoryOne>()));
...
IoCContainer.Register(Component.For<IRepositoryN>().ImplementedBy<RepositoryN>().LifeStyle.Singleton.UsingFactoryMethod(() => Factory.New<RepositoryN>()));
But how can I use UsingFactoryMethod() together with the code from the first example?
TIA
You have to use a bit of reflection to use it, since you don't have the exact type of the object you'd be resolving.
var factoryMethod = typeof(Factory).GetMethod("New", BindingFlags.Static|BindingFlags.Public);
container.Register(
AllTypes.FromAssemblyContaining<RepositoryOne>)
.BasedOn(typeof(IRepository<>))
.WithService.AllInterfaces()
.Configure(x => x.UsingFactoryMethod((k, c) => factoryMethod.MakeGenericMethod(c.RequestedType).Invoke(null, null)));
Related
I have the following code:
KiwiContainer()
..registerFactory((c) => GetOrderListOfItemsUseCase(
repository: c<ShopRepository>()))
The problem here is that the GetOrderListOfItemsUseCase requires an instance of IProfileRepository which the ShopProfileRepository extends from.
NOTE: IProfleRepository is extended by IUserProfileRepository and IShopProfileRepository.
UPDATE: I have the following factories:
..registerFactory<IShopRepository>((c) => c<ShopRepository>())
..registerFactory<IUserProfileRepository>((c) => c<UserProfileRepository>())
The thing is that I want to dynamically inject the proper dependency for GetUserLocationsUseCase.
..registerFactory(
(c) => GetUserLocationsUseCase(repository: c<IProfileRepository>()))
how can I inject in the code the proper subtype of IProfileRepository whether it IShopRepository or IUserProfileRepository?
If you want to dinamically inject either UserProfileRepository or ShopProfileRepository you should register the one you are going to use via its supertype.
KiwiContainer()
..registerFactory<IProfleRepository>((c) => ShopProfileRepository())
..registerFactory((c) => GetOrderListOfItemsUseCase(
repository: c<IProfleRepository>()))
And obtain it via its supertype too, like in the snippet.
I am trying to have a REST design, but I am running in a bit of a problem. I have a resource schedule. Therefore, the normal notation of /schedules/{id} is not very applicable since I would like to have /schedules/{day}/{month}/{year} and then apply REST, and have /edit and such.
Is there a way to do this with Route::resource() ? or do I need to do them through Route::get() ?
As far as I know route::resource only gives you the routes that are detailed in the documentation so for what you want you would need to declare your own route. It is still restful and if it is only one of the resourceful routes you want to change you should still be able to do the following because the routes are prioritized in the order they are declared.
Route::get('schedule/{day}/{month}/{year}/edit', array('as' => 'editSchedule', 'uses' => 'ScheduleController#edit'));
Route::resource('schedule', 'ScheduleController');
Yes, there is a very simple way. Here is an example:
Specify your route like this:
Route::resource("schedules/day.month.year", "ScheduleController");
The request will be like this:
/schedules/day/1/month/12/year/2014
And now you can get all three parameters in show method of your
contoller:
public function show($day, $month, $year)
Hi there this might be handy if you want to call your route by name. Also you can use one or multiple parameters. It works with me on laravel 5.1
According to the laravel docs:
http://laravel.com/docs/5.1/routing#named-routes
Route::get('user/{id}/profile', ['as' => 'profile', function ($id) {
//
}]);
$url = route('profile', ['id' => 1]);
This works with Route:resource aswell.
for example:
Route::resource('{foo}/{bar}/dashboard', 'YourController');
Will create named routes like: {foo}.{bar}.dashboard.show
To call this with the route method, you set it up as followed.
route('{foo}.{bar}.dashboard.show', ['foo' => 1, 'bar'=> 2])
Which will create the url yourdomain.com/1/2/dashboard
Ill hope this is usefull.
Pascal
I am building an MVC application and have the need to send a different cachecontext (wraps info on which cache to use) into my repositories using the DependencyResolver. There are 10 different types of cachecontexts that can be passed in. At the moment, I am using 10 different registrations for 10 different types of interfaces (ICacheContexts). This works but does not seem like the correct way of handling this circumstance. Is there a way to create a factory that knows the object type it is being passed to so I can just determine the correct CacheContext to return and thus only have one registration to maintain?
Current code:
builder.Register(c => new CacheContext(AppConfig.AppBucketName, AppConfig.AppBucketpassword))
.As<ICacheContextForApps>();
builder.Register(c => new CacheContext(AppConfig.AcctBucketName, AppConfig.AcctBucketpassword))
.As<ICacheContextForAccounts>();
etc..
You need to do this from the repository side - for each repository, configure its parameters to resolve the right cache context. E.g. (paraphrasing)
builder.Register(c => new CacheContext(...for appls...))
.Named<ICacheContext>("apps");
builder.RegisterType<AppsRepository>()
.WithParameter((pi, c) => pi.ParameterType == typeof(ICacheContext),
(pi, c) => c.ResolveNamed<ICacheContext>("apps"));
Hope this gets you on the right track.
I'm trying to implement a REST API to my website.
My problem is that the default Zend routing gets in the way. I've first tried using Zend_Rest_Route but I haven't been able to understand how I was supposed to use it correctly for "deep" routes, aka website/api/resource1/filter/resource2/id.
Using the default Zend routing, I'd need to create a gigantic Resource1Controller to take care of all the possible actions, and I don't think it's the "good" way to do this.
I've tried using Resauce ( http://github.com/mikekelly/Resauce/), creating an api module and adding routes, but I'm not able to get it working correctly :
The patterns I added were :
$this->addResauceRoutes(array(
'api/resource' => 'resource',
'api/resource/:id' => 'custom',
'api/resource/filter' => 'resource-filter',
'api/resource/filter/:id' => 'custom',
));
Which then leads to this :
public function addResauceRoutes($routes) {
$router = Zend_Controller_Front::getInstance()->getRouter();
foreach ($routes as $pattern => $controller) {
$router->addRoute($controller,
new Zend_Controller_Router_Route($pattern, array(
'module' => 'api',
'controller' => $controller
)
)
);
}
Zend_Controller_Front::getInstance()->setRouter($router);
website/api/resource gets me the
Resource1Controller, ok
website/api/resource/filter gets me to the
resource1filterController, ok
website/api/resource/filter/:id gets me to
a custom controller, ok
I'd like for website/api/resource/:id to get me to the same custom controller... But it redirects me to the Resource1Controller.
What solution is there for me to correctly create my API ? Is there a good way to do this with Zend_Rest_Route ?
Edit : Mike,
I felt that it was not appropriate for me to use different controllers since I need the pathes "website/api/resource/:id" and "website/api/resource/filter/:id" to give me almost the exact same result (the only difference is that because the filter is there, I may get a message telling "content filtered" here).
I thought it was a waste creating another almost identical controller when I could've used the same controller and just checked if a parameter "filter" was present.
However, I don't want to use the basic Zend routing since for the path "website/api/resource/filter/resource2" I'd like to have a totally different comportment, so I'd like to use another controller, especially since I'm trying to use Zend_Rest_Action and need my controllers to use the basic actions getAction(), putAction(), postAction() and deleteAction().
Please could you explain why it is you need two URI patterns pointing to the same controller. A better solution might be to use a separate controller for each of the two patterns and move any shared logic into your model.
Forcing a unique controller for each routing pattern was an intentional design decision, so I'd be interested to hear more detail about your use case where you feel this isn't appropriate.
I thought it was a waste creating
another almost identical controller
when I could've used the same
controller and just checked if a
parameter "filter" was present.
Personally, I think it is cleaner to move the shared logic into the model and to keep your controllers skinny. To me it's not wasteful, it's just more organised - it will make your code easier to manage over time.
If you really need to use the same controller you could always use a query parameter instead, that would work fine:
api/resource/foo?filter=true
That URI would be taken care of by the first route ('api/resource/:id' => 'custom') for free.
But please consider using two controllers, I think that is a better approach.
Okay, the reason I didn't get the good controllers was because Resauce uses the controller name as the name of the route, which has to be unique - so second url pointing to "custom" controller couldn't work. Now I'm able to get the files I want :)
So instead of what was previously noted, I use directly the $router->addRoute(); and define new names each times, even if pointing to the same controller.
Example :
$router->addRoute('resource', new Zend_Controller_Router_Route('/api/resources/:id', array('module' => 'api', 'controller' => 'resource')));
$router->addRoute('resourceFiltered', new Zend_Controller_Router_Route('/api/resources/filter1/:id', array('module' => 'api', 'controller' => 'resource', 'filter' => 'filter1'));
I'm trying to use Class::DBI with a simple one parent -> may chidren relationships:
Data::Company->table('Companies');
Data::Company->columns(All => qw/CompanyId Name Url/);
Data::Company->has_many(offers => 'Data::Offer'=>'CompanyId'); # =>'CompanyId'
and
Data::Offer->table('Offers');
Data::Offer->columns(All => qw/OfferId CompanyId MonthlyPrice/);
Data::Offer->has_a(company => 'Data::Company'=>'CompanyId');
I try to add a new record:
my $company = Data::Company->insert({ Name => 'Test', Url => 'http://url' });
my $offer = $company->add_to_offers({ MonthlyPrice => 100 });
But I get:
Can't locate object method "add_to_offers" via package "Data::Company"
I looked at the classical Music::CD example, but I cannot figure out what I am doing wrong.
I agree with Manni, if your package declarations are in the same file, then you need to have the class with the has_a() relationship defined first. Otherwise, if they are in different source files, then the documentation states:
Class::DBI should usually be able to
do the right things, as long as all
classes inherit Class::DBI before
'use'ing any other classes.
As to the three-argument form, you are doing it properly. The third arg for has_many() is the column in the foreign class which is a foreign key to this class. That is, Offer has a CompanyId which points to Company's CompanyId.
Thank you
Well, the issue was actually not my code, but my set up. I realized that this morning after powering on my computer:
* Apache + mod_perl on the server
* SMB mount
When I made changes to several files, not all changes seems to be loaded by mod_perl. Restarting Apache solves the issue. I've actually seen this kind of issue in the past where the client and SMB server's time are out of sync.
The code above works fine with 1 file for each module.
Thank you
I really haven't got much experience with Class:DBI, but I'll give this a shot anyway:
The documentation states that: "the class with the has_a() must be defined earlier than the class with the has_many()".
I cannot find any reference to the way you are using has_a and has_many with three arguments which is always 'CompanyId' in your case.