Show a view in laravel? - laravel

In my controller I have:
public function showIndex()
{
return View::make('index');
}
In my views folder I have index.php.
But it's not showing? Where am I going wrong?
I do not wish to use blade templating.

I also had an index.blade.php file in there. It was loading this by default. Deleted the file and now works.

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Laravel routes with vue/vue router

I'm basically using VueRouter to create all my routes which is working really well. However, my application is built using Laravel. So if I refresh any of the routes I get an error that the route is not defined. So at the minute in every controller I've had to add an index function. This just returns the view of my app.blade which is just the usual tags etc and the to load my single page app. I'm just wondering if there is a cleaner solution? I guess I could move the return view into a single Controller and make my Controllers extend this. But I'm just wondering if there is a better way I'm missing?
i.e. one of my routes via VueRouter:
{
path: "/clients",
name: "clients",
component: () => import(/* webpackChunkName: "clients" */ "../resources/js/components/Views/Clients/Clients.vue")
},
The route in my clients.php file
Route::get('/clients', [App\Http\Controllers\ClientController::class, 'index'])->name('clients');
Then the ClientController index function
public function index()
{
return view('app');
}
It would just be nice to have the loading of the app.blade done somewhere else and not need to be specified per controller. Any help would be appreciated to ensure it's all done efficiently!
Thanks!
Here is how I solved this issue for one of my projects which is also single page application in Vue and Laravel: https://github.com/lyyka/laravel-vue-blog-spa/blob/master/routes/web.php
Simply, in your routes file, you put this code:
Route::get('/{any}', function () {
return view('welcome');
})->where("any", ".*");
And in my welcome view I have:
#extends('layouts.app')
#section('content')
<div class = "container">
<router-view></router-view>
</div>
#endsection
Basically this will return your app view for any URL and so your Vue SPA should work properly. Generally, it is not a good practice to put callback functions inside your routes file, but in this case, you won't even be using your routes file as it is a SPA, so this solution can pass! :)
You should use your single html file and make a controller.
On your controller
public function index(){
return view('index');
}
on your web.php
basically, you should make the same route on your laravel and vue
Route::get('/products', [ProductsController::class,'index']);
in my vue-routes
import Products from './components/Products.vue'
{
path:'/products'
component: Products
}

while page yet, i have set everything

I am trying to make via hi in view but its give me white page display. I have created controller and view ... path is also set in web.php
ROUTE:
Route::post('/member/add-single-trade/import_single_trades_parse', 'trades\ImportSingleTradesController#tradesImport');
CONTROLLER:
public function tradesImport()
{
return view('member.add-single-trade.import-excel.import_fields', compact( ''));
}
BLADE: import_fields.blade.php
hi
why dont you just return 'hi' in tradesImport() to see whether the route works fine
SOLVED: underscore does not work in route: /member/add-single-trade/import_single_trades_parse
It's working with : /member/add-single-trade/import-single-trades-parse

Laravel disable controller action layout

Is there a way to disable layout for certain controller method?
Im using something like $this->layout = null ,yet it still render the layout
The view im rendering obviously have a layout associate with it, i just wonder is it possbile to disable the layout from within controller method, without need to modify the blade file itself
Here is the controller:
class PurchaserController extends \BaseController
{
public function index()
{
$this->layout = null;
return View::make('purchasers.index');
}
}
The view:
#extends('layouts.master')
#section('content')
Content
#stop
Im using Laravel 4
Just remove
#extends('layouts.master')
from your view. That will prevent the view from loading.
Also - if you are using the #extends - then you dont actually need $this->layout() in your controller at all
Edit:
" i just wonder is it possbile to disable the layout from within controller method, without need to modify the blade file itself"
The idea is you do it either entirely from the controller, or entirely from the blade file. Not both together.

How to call data and view it on the footer?

This is the function that I am using in my controller
public function homeList()
{
//Get all the franchises
$franchises = Franchise::all();
//Load the view and pass the franchises
return View::make('frontend.layouts.footer')->with('franchises', $franchises);
}
and I keep getting this error. I don't know how to pass it or what to put on the routes.php file
You should be able to do #include('frontend.layouts.footer')->with('franchises', Franchise::all()).
Update:
To avoid the model in the view, you should use a view composer, as it was stated in a previous answer.
View::composer('frontend.layouts.footer', function($view)
{
$view->with('franchises', Franchise::all());
});
Now you have $franchises available in the view. This code you can place in routes.php or you can create a composers.php and autoload it.

Laravel 4: if statement in blade layout works strange

Could someone explain me why I get blank screen with printed string "#extends('layouts.default')" if I request page normally (not ajax)?
#if(!Request::ajax())
#extends('layouts.default')
#section('content')
#endif
Test
#if(!Request::ajax())
#stop
#endif
I'm trying to solve problem with Ajax, I don't want to create 2 templates for each request type and also I do want to use blade templates, so using controller layouts doesn't work for me. How can I do it in blade template? I was looking at this Laravel: how to render only one section of a template?
By the way. If I request it with ajax it works like it should.
Yes #extends has to be on line 1.
And I found solution for PJAX. At the beginning I was not sure this could solve my problem but it did. Don't know why I was afraid to lose blade functionality if you actually can't lose it this way. If someone is using PJAX and needs to use one template with and without layout this could be your solution:
protected $layout = 'layouts.default';
public function index()
{
if(Request::header('X-PJAX'))
{
return $view = View::make('home.index')
->with('title', 'index');
}
else
{
$this->layout->title = 'index';
$this->layout->content = View::make('home.index');
}
}
Try moving #extends to line 1 and you will see the blade template will render properly.
As for solving the ajax problem, I think it's better if you move the logic back to your controller.
Example:
…
if ( Request::ajax() )
{
return Response::eloquent($books);
} else {
return View::make('book.index')->with('books', $books);
}
…
Take a look at this thread for more info: http://forums.laravel.io/viewtopic.php?id=2508
You can still run your condition short handed in the fist line like so
#extends((Request::ajax())?"layout1":"layout2")

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