I have a link on a page /my/example/page which links to a laravel route my.route.
Route
Route::group(['middleware' => ['auth']], function () {
Route::group(['middleware' => ['Some\Custom\Auth\Middleware:1']], function () {
Route::match(['GET', 'POST'], '/my/route/to/external/controller', 'exampleController#externalLink')->name('my.route');
}
}
Link on /my/example/page
Link
This route /my/route/to/external/controller points to this controller method externalLink in the exampleController controller, which returns the url for the href to use
public function externalLink()
{
return $this->redirect->away('www.externalsite.com');
}
My test is
$this->visit(/my/example/page)
->click('Link')
->assertRedirectedToRoute('my.route');
I constantly get the error
Symfony\Component\HttpKernel\Exception\NotFoundHttpException
when I use the click() testing method.
I can catch this using #expectedException but his doesn't help as I am expecting to see a different page.
I have also tried (not together);
->assertResponseStatus(200);
->seePageIs('www.externalsite.com');
->assertRedirect();
->followRedirects();
From browser inspection, when the url is clicked I get
http://www.example.com/my/route/to/external 302
http://www.externalsite.com 200
How can I functionally test the button being clicked and redirecting to an external site?
I am struggling with a similar problem right now, and I have just about given up on testing the endpoint directly. Opting for a solution like this...
Test that the link contains proper information in the view:
$this->visit('/my/example/page')
->seeElement('a', ['href' => route('my.route')]);
Moving the logic in the controller to something you can test directly. The Laravel\Socialite package has some interesting tests that might be helpful if you do this...
class ExternalLinkRedirect {
public function __construct($request){
$this->request = $request;
}
public function redirect()
{
return redirect()->away('exteranlsite.com');
}
}
Then test it directly
$route = route('my.route');
$request = \Illuminate\Http\Request::create($route);
$redirector = new ExternalLinkRedirect($request);
$response = $redirector->redirect();
$this->assertEquals('www.externalsite.com', $response->getTargetUrl());
A simple way is to use the get method instead of visit.
An example:
$this->get('/facebook')
->assertRedirectedTo('https://www.facebook.com/Les-Teachers-du-NET-516952535045054');
Related
Can anyone tell why the documentation of Laravel, and others, show functions in routes that return / do something? In what context can you use this?
For example, I try to figure out Molly Connect.
Here is the corresponding code from https://github.com/mollie/laravel-mollie/blob/master/docs/mollie_connect.md
Route::get('login', function () {
return Socialite::with('mollie')
->scopes(['profiles.read']) // Additional permission: profiles.read
->redirect();
});
Route::get('login_callback', function () {
$user = Socialite::with('mollie')->user();
Mollie::api()->setAccessToken($user->token);
return Mollie::api()->profiles()->page(); // Retrieve payment profiles available on the obtained Mollie account
});
Its just a shortcut, to avoid having to create separate controller files and indirectly referencing those functions. Functionally, your example is no different from doing this:
Route::get('login_callback', 'LoginController#callback')
And then, LoginController.php
class LoginController
{
public function callback()
{
$user = Socialite::with('mollie')->user();
Mollie::api()->setAccessToken($user->token);
return Mollie::api()->profiles()->page();
}
}
See here
I searched more time to find how to get the controller method name by passing the URL but not found my expected answer. I want to make a method where I will pass a URL and it will give the corresponding controller action like as below but I can't figure out.
I found a helper which just return the current URL's action which is Route::currentRouteAction()
If a route in my application like as Route::get('/abc', 'YourController#method') which will generate the url http://example.com/abc
then how can I get the YourController#method by passing http://example.com/abc
function getAction($url){
//what will be logic?
// return like App\Controllers\MyController#method
}
I have to make a custom permission system where I need it for show and hide the menu by checking the URL of each menu.
Within your controller you can do the following:
<?php
use Illuminate\Routing\Router;
use Illuminate\Http\Request;
public function index(Request $request, Router $route)
{
$action = $router->getRoutes()->match($request)->getActionName();
// action should be what you're looking for.
}
You can try this if you want to:
Route::get('/the/url', 'YourController#method');
Every time anything calls the URL in the route, your method will be called.
You don't need to navigate to that url to call your method, it could be called by a form action, or a buttons action and just execute your method.
Edit:
url is your url as parameter (plain route)
import this:
use Illuminate\Routing\Route;
this is your function:
public function method(Route $route, $url)
{
$routes = \Route::getRoutes()->getRoutes();
foreach($routes as $r){
if($r->getUri() == $url){
$youraction= $r->getActionName();
dd($youraction);
}
else{
dd('does not exist');
}
}
}
Tested.
im new in laravel 5.2, I just want to ask how you can catch a link that came from uploads like: http://sitename.com/uploads/59128.txt? I want to redirect them to login page if they tried to access any of route or link that came from uploads/{any filename}.
Yes you can achieve by protecting your route with auth middleware,
make a small FileController
class FileController extends Controller {
public function __construct()
{
$this->middleware('auth');
}
public function getFile($filename)
{
return response()->download(storage_path($filename), null, [], null);
}
}
and then in routes.php
Route::get('file/{filename}', 'FileController#getFile')->where('filename', '^[^/]+$');
And that's it. Now, your authenticated users can download files from storage folder (but not its subfolders) by calling http://yoursite.com/file/secret.jpg. Add you can use this URL in src attribute of an image tag.
answer's original source!
#xerwudjohn simple you can't.
When this file is in the public folder, everyone can access it whitout being logged in.
One method I tried for some minutes, create a new route:
Route::group(['middleware' => ['web', 'auth']], function () {
Route::get('/download/{id}', 'DownloadController#showFile');
});
create the function showFile in the DonwloadController
public function showFile($id)
{
return redirect('/image/'.$id.'.txt');
}
or use a Model to read uniqueIds out of any table and get the realfile name.
Cheers
I am trying to reopen same page on Codeigniter. See code below:
<?php
class Leads extends CI_Controller
{
public $tempName;
public $tempFrom;
public $tempTo;
public function index() {
$this->load->model('Lead');
$leads = $this->Lead->getAllLeads();
$pageNum = count($leads);
$this->load->view('main', array('leads' => $leads, 'pageNum' => $pageNum));
}
public function getLeads(){
$this->load->model('Lead');
$name = $this->input->post('name');
$from = $this->input->post('from');
$to = $this->input->post('to');
$leads = $this->Lead->getLeads($name, $from, $to);
$pageNum = count($leads);
$this->load->view('main', array('leads' => $leads, 'pageNum' => $pageNum));
}
}
?>
As you can see, first I open 'main' from 'index()'. In main, I call 'getLeads()', but my page is not getting updated. Is it because I can't refresh data while on the same page? If yes, how can I work around it?
Thanks a lot!
You cannot call a controller from a view. It sounds like you probably should be using AJAX in your "main" view to get additional data from getLeads().
Option A:
redirect with js to the same function you have, but change the method to get (send params in the URL and get them on the server with $this->input->get()).
Option B:
Since you are posting to getLeads I assume you are using ajax. And since you are using ajax, you should return an url or data to render on your website. Maybe you could use $this->session->set_flashdata('leads',$leads) and $this->session->set_flashdata('pageNum',$pageNum) , send an url to another controller function, extract leads and pageNum $this->session->flashdata('leads') & $this->session->flashdata('pageNum') and then render the view?
Responding with load view will not automatically set the html to what you get.
I'm trying to generate ajax specific responses from my controllers by using the Request::ajax() method, which is working just fine. The only problem is that the way I have it set up right now isn't really a nice looking solution.
My controller:
class HomeController extends BaseController {
protected $layout = 'layouts/main';
public function __construct()
{
$this->beforeFilter('auth');
}
public function getIndex()
{
$view = View::make('content.home.index');
if(Request::ajax()) return $view; //For ajax calls we only want to return the content to be placed inside our container, without the layout
$this->layout->menu = 'content.menu';
$this->layout->content = $view;
}
}
So right now, for every method I define within my controllers I need to add the code snippet that checks for an AJAX request and returns a single view if the statement returns true.
This leads to my question that is probably more PHP related than it is to the framework;
Is there a way of executing my AJAX check on every method call, without actually placing it inside the method? Or is there some other solution to keep my code DRY?
Thanks in advance!
PS: This is my first post on stackoverflow, so feel free to correct me if I made any mistakes
Create a new barebone layout named 'layouts/ajax' (or any name you like).
<?php echo $content ?>
In your Base controller, override this setupLayout() function.
protected function setupLayout()
{
if ( ! is_null($this->layout))
{
$layout = Request::ajax() ? 'layouts/ajax' : $this->layout;
$this->layout = View::make($layout);
}
}
Change your getIndex() function to this.
public function getIndex()
{
$view = View::make('content.home.index');
$this->layout->menu = 'content.menu';
$this->layout->content = $view;
}
Now non-ajax requests will be rendered using layout set in the controller, where as ajax requests will receive whatever set to $this->layout->content.
Note : Controller will neglect the layout setup in setupLayout(), if the called method returns truthy value. So this method will not work for functions like below.
public function getIndex()
{
return View::make('content.home.index');
}
You could just change the layout property, in the constructor, if it's an ajax request:
public function __construct()
{
$this->beforeFilter('auth');
if(Request::ajax()) {
$this->layout = '';
}
}
If it doesn't work try setting it to NULL instead.
Why would you return a VIEW via ajax? Are you using it to create a SPA? If so there are better ways. I'm generally against returning HTML via AJAX.
The route I'd go in your position is probably opposite of how you're doing it. Render the view no matter what, if the request is ajax, pass the extra data back and have JS render the data on the page. That's essentially how most Javascript MVC frameworks function.
Sorry if I am totally missing the point here, just going on an assumption of your end goal with the info you provided.