I already implement this code and it works,
function to return/generate html from view file, i add the view() method to the controller
//i'm ok with this and did works
Route::get('/page/{name}', function($name){
if(class_exists("App\Http\Controllers\\".$name."Ctrl")){
return App::make("App\Http\Controllers\\".$name."Ctrl")->view();
}
else{
return "Page tidak ditemukan";
}
});
Now i want to implement this on resource route, but cannot find technic/method to function the same way as above
Route::resource('/data/{name}', function($name){
//check first if controller exists
if(class_exists("App\Http\Controllers\\".$name."Ctrl")){
//need some Kind Of method or hack i presume
// so the request is processed as defined by the standar resource route
//
//NEED HELP ON THIS FUNCTIONALITY
//
}
else{
return "Page tidak ditemukan";
}
});
Related
I have a named route with a parameter which looks like this...
Route::get('/profile/{user_id}', [ProfileController::class, 'profile'])->name('profile');
Now in one of my controller,
I have a function that calls this route like this
public function _myFunction($some_data) {
return redirect()->route('profile', [$some_data->user_id]);
}
and in my ProfileController's profile() function, I have this.
public function profile() {
return view('modules.profile.profile');
}
I've followed the documentation and some guides I found in SO, but I got the same error,
"Route [profile] not defined."
can somebody enlighten me on where I went wrong?
Here's what my routes/web.php looks like...
Route::middleware(['auth:web'])->group(function ($router) {
Route::get('/profile/{user_id}', [ProfileController::class, 'profile'])->name('profile');
});
When calling the route, you should pass the name of the attribute along with the value (as key vaue pairs). In this case, your route is expecting user_id so your route generation should look like this:
return redirect()->route('profile', ['user_id' => $some_data->user_id]);
Read more on Generating URLs To Named Routes in Laravel.
I solved the issue, and its really my bad for not providing a more specific case information and made you guys confused.
I was using socialite and called _myFunction() inside the third party's callback..
After all, the problem was the socialite's google callback, instead of placing the return redirect()->route('profile', [$user->id]) inside _myFunction(), what I did was transfer it to the callback function.
So it looked like this now...
private $to_pass_user_id;
public function handleGoogleCallback()
{
try {
$user = Socialite::driver('google')->user();
$this->_myFunction($user);
return redirect()->route('profile', [$this->to_pass_user_id]);
} catch (Exception $e) {
dd($e->getMessage());
}
}
public function _myFunction($some_data) {
... my logic here
$this->to_pass_user_id = $some_value_from_the_logic
}
I'm following a tutorial on Laravel, adding to a DB via a form. At the end of a function that saves to a DB, it returns back to the page where the form is, but I want to be taken to another page where the information is displayed. In the tutorial I created a controller with a function that returns a view containing all the database info - that element works fine however I can't seem to find a way of calling this function directly after saving to the database. I can also return any other view which just displays static view ( just html with no data handling ). Is what I'm trying to achieve possible?
public function store(){
$li = new \App\LTest1();
$li->creator = request('creator');
$li->title = request('title');
$li->views = request('views');
$li->save();
return back(); // this works
// return view('info'); // this works
//return ('Listings#showList'); this doesnt work = how do i call a function in a controller???
}
// routing
Route::get('info', function () {
return view('info'); // i can get to this static page from my store() function
});
Route::get('thedataviewpage', 'Listings#showList'); // you can route to this but not from the store() function
Redirect is the thing you need here
public function store() {
$li = new \App\LTest1();
$li->creator = request('creator');
$li->title = request('title');
$li->views = request('views');
$li->save();
return redirect('info'); // Redirect to the info route
}
Take this example. Be sure to add the proper route name and a proper message.
return redirect()->route('put here the route name')->with('success', 'Created.');'
to return to a controller action just use
return redirect()->action('Listings#showList');
or you can use route to call that controller action
return redirect('/thedataviewpage');
I want to check segment of particular URL on route and based on value of segment decide it to handle to another route.Somewhat like below:
Route::get('{module}/{seg}', function(){
if (is_numeric((Request::segment(3)) {
return Route::get('{module}/{seg}',Request::segment(2) . 'Controller#index');
}else{
return Route::get('{module}/{seg}',Request::segment(2).'Controller#index' . Request::segment(3));
}
});
I don't think above code works but can anyone suggest a working code for implementing above logic in laravel?
I'd suggest adding it as an optional parameter, and handle differences in the controller. Given your code, it might look like this, for instance:
// route
Route::get('{module}/{seg}/{param?}', 'Controller#index');
// controller
public function index($module, $seg, $param = null)
{
// for dynamic index methods
if (is_numeric($param)) {
$method = 'index' . $param;
return $this->{$method}();
}
// for non-numeric third-segment params, continue here as usual
}
How do I create custom view for errors on Lumen? I tried to create resources/views/errors/404.blade.php, like what we can do in Laravel 5, but it doesn't work.
Errors are handled within App\Exceptions\Handler. To display a 404 page change the render() method to this:
public function render($request, Exception $e)
{
if($e instanceof NotFoundHttpException){
return response(view('errors.404'), 404);
}
return parent::render($request, $e);
}
And add this in the top of the Handler.php file:
use Symfony\Component\HttpKernel\Exception\NotFoundHttpException;
Edit: As #YiJiang points out, the response should not only return the 404 view but also contain the correct status code. Therefore view() should be wrapped in a response() call passing in 404 as status code. Like in the edited code above.
The answer by lukasgeiter is almost correct, but the response made with the view function will always carry the 200 HTTP status code, which is problematic for crawlers or any user agent that relies on it.
The Lumen documentation tries to address this, but the code given does not work because it is copied from Laravel's, and Lumen's stripped down version of the ResponseFactory class is missing the view method.
This is the code which I'm currently using.
use Symfony\Component\HttpKernel\Exception\HttpException;
[...]
public function render($request, Exception $e)
{
if ($e instanceof HttpException) {
$status = $e->getStatusCode();
if (view()->exists("errors.$status")) {
return response(view("errors.$status"), $status);
}
}
if (env('APP_DEBUG')) {
return parent::render($request, $e);
} else {
return response(view("errors.500"), 500);
}
}
This assumes you have your errors stored in the errors directory under your views.
It didn't work for me, but I got it working with:
if($e instanceof \Symfony\Component\HttpKernel\Exception\NotFoundHttpException) {
return view('errors.404');
}
You might also want to add
http_response_code(404)
to tell the search engines about the status of the page.
I am using the HMVC pattern in CodeIgniter. I have a controller that loads 2 modules via modules::run() function and a parameter is passed to each module.
If either module cannot match the passed paramter I want to call show_404(). It works, but it loads the full HTML of the error page within my existing template so the HTML breaks and looks terrible. I think I want it to redirect to the error page so it doesn't run the 2nd module. Is there some way to do that and not change the URL?
Is it possible to just redirect to show_404() from the module without changing the URL?
Here is an over simplified example of what's going on:
www.example.com/users/profile/usernamehere
The url calls this function in the users controller:
function profile($username)
{
echo modules::run('user/details', $username);
echo modules::run('user/friends', $username);
}
Which run these modules, which find out if user exists or not:
function details($username)
{
$details = lookup_user($username);
if($details == 'User Not Found')
show_404(); // This seems to display within the template when what I want is for it to redirect
else
$this->load->view('details', $details);
}
function friends($username)
{
$details = lookup_user($username);
if($friends == 'User Not Found')
show_404(); // Redundant, I know, just added for this example
else
$this->load->view('friends', $friends);
}
I imagine there is just a better way to go at it, but I am not seeing it. Any ideas?
You could throw an exception if there was an error in a submodule and catch this in your controller where you would do show_404() then.
Controller:
function profile($username)
{
try{
$out = modules::run('user/details', $username);
$out .= modules::run('user/friends', $username);
echo $out;
}
catch(Exception $e){
show_404();
}
}
Submodule:
function details($username)
{
$details = lookup_user($username);
if($details == 'User Not Found')
throw new Exception();
else
// Added third parameter as true to be able to return the data, instead of outputting it directly.
return $this->load->view('details', $details,true);
}
function friends($username)
{
$details = lookup_user($username);
if($friends == 'User Not Found')
throw new Exception();
else
return $this->load->view('friends', $friends,true);
}
You can use this function to redirect 404 not found page.
if ( ! file_exists('application/search/'.$page.'.php')) {
show_404(); // redirect to 404 page
}
its very simple , i solved the problem
please controler name's first letter must be capital e.g
A controller with
pages should be Pages
and also save cotroler file with same name Pages.php not pages.php
also same for model class
enjoy