Can I create a Controller that simply returns an image?
I would like to route this logic through a controller, whenever a variable [LOGO] such requested,
The controller will look up logo.png and replace the image to the variable [LOGO].
I don't want to use a View. I want to do it all with just the Controller.
Is this possible?
I tried some way by following code,
EmailHelper.SendEmail(
Constants.EmailSender.CandidateSupport,
candidate.Email,
Constants.EmailSubject.CandidateUpdateProfile,
Constants.EmailBody.CandidateUpdateProfile
.Replace("[LOGO]", Url.Content("~/Content/Images/logo_small2.png"))
I am not entirely sure what you need, but in answer to "Can I create a Controller that simply returns an image?" you can do this as follows:
public FileResult Image(string filePath)
{
//get your stream
return File(stream, "image/png");
}
Obviously you'd need to think about security given that you are returning items from the file system.
Related
So basically, I have a setup of restful controller in my route. Now my problem is how can I call the Index page if there is a parameter.. it gives me an error of Controller not found
Im trying to call it like this www.domain.com/sign-up/asdasdasd
Route::controller('sign-up','UserRegisterController');
then in my Controller
class UserRegisterController extends \BaseController {
protected $layout = 'layouts.unregistered';
public function getIndex( $unique_code = null )
{
$title = 'Register';
$this->layout->content = View::make( 'pages.unregistred.sign-up', compact('title', 'affiliate_ash'));
}
By registering:
Route::controller('sign-up','UserRegisterController');
You're telling the routes that every time the url starts with /sign-up/ it should look for corresponding action in UserRegisterController in verbAction convention.
Suppose you have:
http://domain.com/sign-up/social-signup
Logically it'll be mapped to UserRegister#getSocialSignup (GET verb because it is a GET request). And if there is nothing after /sign-up/ it'll look for getIndex() by default.
Now, consider your example:
http://domain.com/sign-up/asdasdasd
By the same logic, it'll try looking for UserRegister#getAsdasdasd which most likely you don't have. The problem here is there is no way of telling Route that asdasdasd is actually a parameter. At least, not with a single Route definition.
You'll have to define another route, perhaps after your Route::controller
Route::controller('sign-up','UserRegisterController');
// If above fail to find correct controller method, check the next line.
Route::get('sign-up/{param}', 'UserRegisterController#getIndex');
You need to define the parameter in the route Route::controller('sign-up/{unique_code?}','UserRegisterController');. The question mark makes it optional.
Full documentation here: http://laravel.com/docs/routing#route-parameters
I have some parameters in URL, which I would like to be present in the URL for all pages in my MVC3 app. For example:
mycompany.com/home?param=1
mycompany.com/cart?param=1
mycompany.com/logout?param=1
Whether the user is navigating to a new page or submitting a form, how can I have my parameter
be present in all my pages? Right now the only way I can think off is somehow reconstruct the URL for every new view I need to render. Is there built in functionality in MVC to do this?
Thanks
That sounds like something you should store in Session instead of updating all of your links to add the same parameter.
You could use a Session variable as the other poster suggested, or you could use a base view model which holds this static param value.
So your base view would be something like this:
public class BaseViewModel
{
public static int ParamValue = 1;
}
then in each view model you use for each view, you'd have something like this:
public class PageViewModel : BaseViewModel
{
// properties
}
This way, in each view, you can just reference #Model.ParamValue whenever you need to access it:
#Model Namespace.PageViewModel
My param value is <b>#Model.ParamValue</b>
I'm trying to implement a common controller in MVC3 to return various JSON feeds, example -
public class AjaxController : Controller
{
public ActionResult Feed1()
{
ViewBag.Json = LogicFacade.GetFeed1Json();
return View();
}
public ActionResult Feed2()
{
ViewBag.Json = LogicFacade.GetFeed2Json();
return View();
}
}
This class has 30+ methods in it, the problem is this requires implementing an IDENTICAL View for each of the Controller's methods (sigh) that writes out ViewBag.Json.
I'm assuming this is a routing issue but I'm struggling with that. The following didn't work -
Tried setting ViewBag.Json then using RedirectToAction() but that seems to reset ViewBag.Json.
Note JsonResult is not appropriate for my needs, I'm using a different JSON serialiser.
So the objective here is to maintain one View file but keep this class with seperate methods that are called by routing, and not a crappy switch statement implementation.
Any help appreciated.
Use the same view and just specify the name. You can store in the controller's view folder, if only used by one controller, or in the Shared view folder if used by more than one.
return View("SharedJsonView");
Another, perhaps better, solution would be to create your own result -- maybe deriving from JsonResult, maybe directly from ActionResult -- that creates the JSON response that you need. Look at the source code for JsonResult on http://www.codeplex.com/aspnet for ideas on how to do it.
I'm using the MVC PHP framework Codeigniter and I have a straight forward question about where to call redirect() from: Controller or Model?
Scenario:
A user navigates to www.example.com/item/555. In my Model I search the item database for an item with the ID of 555. If I find the item, I'll return the result to my controller. However, if an item is not found, I want to redirect the user somewhere. Should this call to redirect() come from inside the model or the controller? Why?
No your model should return false and you should check in your controller like so:
class SampleModel extends Model
{
//Construct
public function FetchItem($id)
{
$result = $this->db->select("*")->from("table")->where("item_id",$id)->get();
if($result->num_rows() == 0)
{
return false;
}
//return result
}
}
and within your controller do:
function item($id)
{
$Item = $this->SampleModel->FetchItem($id);
if(!$Item)
{
redirect("class/error/no_item");
}
}
Models are for data only either return a standard result such as an key/value object or a boolean.
all logic should be handled / controlled by the Controller.
Models are not page specific, and are used globally throughout the whole application, so if another class / method uses the model, it might get redirect to the incorrect location as its a different part of your site.
It seems like the controller would be the best place to invoke your redirect because the controller typically delegates calls to the model, view, or in your case, another controller.
However, you should use whatever makes the most sense for your application and for what will be easier to maintain in the future, but also consider that rules do exist for a reason.
In short, if a coworker were to try to fix a bug in your code, what would the "reasonable person" standard say? Where would most of them be most likely to look for your redirect?
Plus, you said you're returning the result to your controller already... perhaps that's where you should make your redirect...
I'm trying to make URL-friendly links for the blog on my portfolio.
So I would like to obtain links something like site/journal/post/{title}
Obviously Journal is my controller, but let's say my title would be 'mysite.com goes live!' I would like to have a valid url like site/journal/post/mysitecom-goes-live where all disallowed characters are removed.
How would I transform 'mysite.com goes live!' to 'site/journal/post/mysitecom-goes-live' in CodeIgniter based on the characters in $config['permitted_uri_chars']
use the url helper
$this->load->helper('url');
$blog_slug = url_title('Mysite.com Goes live!');
echo $blog_slug //mysitecom-site-goes-live
// might differ slightly, but it'll do what you want.
to generate url-friendly links.
Store this value in a field in your blog table (url_title/url_slug) whatever.
make a function:
class Journal extends controller
{
//make your index/constructor etc
function view($post)
{
$this->blog_model->get_post($post);
// etc - your model returns the correct post,
// then process that data and pass it to your view
}
}
your blog_model has a method get_post that uses CI's
$this->db->where('url_title', $post);
hope that makes sense.
then when you access the page:
site.com/journal/view/mysite-goes-live
the function will pick up "mysite-goes-live" and pass it to the view() function, which in turn looks up the appropriate blog entry in the database.