In my head section of my form.layout.blade.php I have the following:
<head>
<script src="/js/main.js"></script>
<head>
This is layout file is loaded before all pages are loaded. Is there a way to not load main.js for a specific route?
you can use if statement like this. main.js will not load on this page
#if(!Request::is('subpage/url'))
<script src="/js/main.js"></script>
#endif
The workaround I used for not loading scripts on all sites, is that I put script paths in variables in controllers, something like this:
public function show($id){
$scripts[] = '/js/main.js';
view()->share('scripts', $scripts);
return view('view', $someData);
}
This allows me to use this script only on a page I really need it, like this:
#if(isset($scripts))
#foreach($scripts as $key => $value)
<script src="{{mix($value)}}"></script>
#endforeach
#endif
Hope that this can lead you in right direction.
You can use if statements
#if(!in_array(Route::currentRouteName(), ['route.name', ...]))
<script src="/js/main.js"></script>
#endif
Related
Ths should be simple enough, but...
I have a layout view dashboard-layout.blade.php with:
#section('headscripts')
<script src="{{ mix('/js/app.js') }}" defer></script>
#show
And a billing.blade.php view that extends the layout file:
#extends('layouts.dashboard-layout')
#section('headscripts')
#parent
<script src="https://js.braintreegateway.com/web/dropin/1.17.2/js/dropin.min.js"></script>
#endsection
The problem is, this only outputs the first script...what am I doing wrong?
You can simply use #stack('name here') instead of #yield('name here')
and use
#push('name here')
// Add Your Script Here
#endpush
instead of
#section('name here')
// Scripts
#endsection
to solve your problem.
this is my code
route.php
Route::get('test/{id}','testController#test');
testController.php
public function test($id){ return('test'); }
test.blade.php
<!DOCTYPE html>
<html>
<head>
<title>a</title>
</head>
<body>
<img src="myPath/myImage.jpg">
</body>
</html>
The result: the image cant be load
note: when i deleted the 'id' parameter it has worked perfectly
What is the error message you are getting?
You will also probably want to change the return value of testController#test to something along the lines of this so you can actually pass the variable to the view.
return view('test', [
'id' => $id,
])
You may also want to have env('APP_DEBUG',true) turned on to get verbose output
the solution : i have changed the value of the src attribute
<img src="{{URL::to('myPath/myImage.jpg')}}">
You need to use:
href="{{ URL::to('yourfile')}}". Also you should place your images in laravel public folder. If you will create folder img in your public folder then you need to something like this: href="{{ URL::to('img\imgname.png')}}".
I'm still new to Laravel. Anyway, I'm making a small system of viewing and creating articles to be shown in the main site. I wanted to create a page that displays an article from the database, using a parameter from the URL, like this:
http://localhost:8000/read/1
The above, for example, will display the article with 'id' value of 1 in the database.
So it works just fine, but problem is, after I got it to work (which took me some time since I'm a newbie), the whole style just disappears from the page. I tried to rewrite everything but it still didn't work. New pages that I create include the style just fine.
This is my route line:
Route::get('read/{id}', array('as' => 'read', 'uses' => 'NewsController#readArticle'));
NewsController readArticle function:
public function readArticle($id) {
$article = NewsMessage::where('id', $id) -> first();
return View::make('news.read', array('article' => $article));
}
And the file read.blade.php (located in views/news/read.blade.php)
#extends('layouts.main')
#section('content')
{{ $article -> title }}
#stop
So the whole PHP code works fine and I manage to get the title. But for some reason, the whole style disappears and this is what I see:
http://puu.sh/ctO6J/db30fbe102.png
So any idea, what have I dont wrong that caused this? The other pages work just fine with the style included.
Thank you!
The issue is that the path to the image is incorrectly specified.
You can either use the built in asset methods or something like {{ URL::to('/')/imagepathhere/filename.jpg }}
Assuming you have your styles links wrote as:
#extends('layouts.main')
#section('styles')
<link href="css/custom.css" rel="stylesheet" />
<link href="css/styles.css" rel="stylesheet" />
#endsection
#section('content')
{{ $article -> title }}
#endsection
Correct Syntax:
#extends('layouts.main')
#section('styles')
<link href="{{ asset('css/custom.css') }}" rel="stylesheet" />
<link href="{{ asset('css/styles.css') }}" rel="stylesheet" />
#endsection
#section('content')
{{ $article -> title }}
#endsection
And make sure your css folder is located at public folder when calling with asset() helper.
This issues of disappearing the whole page styles is causing by accessing the image or css file or any other resources, and the solution is to use asset() helper function.
NOTE: YOUR ROUTES AND CONTROLLER HAS NO ISSUE, YOU DON'T NEED TO TOUCH EITHER.
I am using a fresh build today of Laravel 4.
I have a dashboardController
class DashboardController extends BaseController {
protected $layout = 'layouts.dashboard';
public function index()
{
$this->layout->content = View::make('dashboard.default');
}
}
I have a simple route
Route::get('/', 'DashboardController#index');
I have a blade layout in views/layouts/dashboard.blade.php
For the sake of saving everyone from all of the actual HTML ill use a mock up.
<html>
<head>
<title></title>
</head>
<body>
#yield('content')
</body>
</html>
I have a default blade file in views/dashboard/ that has the following (edited for simplicity)
#section('content')
<p>This is not rocket science</p>
#stop
For some reason the content gets generated before the layout.
I am using a different approach to set the layouts globally to routes using a custom filter. Put the following filter into the app/filters.php
Route::filter('theme', function($route, $request, $response, $layout='layouts.default')
{
// Redirects have no content and errors should handle their own layout.
if ($response->getStatusCode() > 300) return;
//get original view object
$view = $response->getOriginalContent();
//we will render the view nested to the layout
$content = View::make($layout)->nest('_content',$view->getName(), $view->getData())->render();
$response->setContent($content);
});
and now instead of setting layout property in the controller class, you can group the routes and apply the filter as shown below.
Route::group(array('after' => 'theme:layouts.dashboard'), function()
{
Route::get('/admin', 'DashboardController#getIndex');
Route::get('/admin/dashboard', function(){ return View::make('dashboard.default'); });
});
When creating the views, make sure to use the #section('sectionName') in all the sub views and use #yield('sectionName') in the layout views.
I find it easier to do my layout like this for example. I would create my master blade file like so
<html>
<body>
#yield('content');
</body>
</html
And in the blade files that I want to use the master at the top i would put
#extends('master')
then content like so
#section('content')
// content
#stop
Hope this helps.
When you use controller layouts, i.e. $this->layout->..., then you get access to data as variables, not sections. So to access content in your layout you should use...
<html>
<head>
<title></title>
</head>
<body>
<?php echo $content; ?>
</body>
</html>
And in your partial, you would not use #section or #stop...
<p>This is not rocket science</p>
My html document looks like this:
<html>
<head> .. load jquery and other stuff </head>
<body>
<div id="cool_container">
<div class="cool">.. no script friendly markup ..</div>
</div>
<a id="cool_link">Link</a>
<script>
function installStuff(){
$('.cool').coolPlugin();
$('#cool_link').click(function(){
$('#cool_container').load('/anothercooldiv.html');
});
}
$(document).load(function(){ installStuff(); });
</script>
</body>
</html>
Of course, /anothercooldiv.html gives another <div class="cool"> .. etc ...</div> fragment.
So what's the best way to turn the fresh cool div into a coolPlugin without breaking everything (and writing some nasty hacks) ?
It'd would be great to be able to either:
Call installStuff with a default jQuery context '#cool_container', so I could call something like:
$.doThisInContext(function(){installStuff();}, $('#cool_container');
In the load callback.
Or, have an equivalent of 'live' (that would solve the problem of links if cool contains links), but on an element existence, that I could use like that in my function installStuff:
$('.cool').exists(function(what){ what.coolPlugin() };
Then the coolPlugin would be installed on all cool elements now and in the future.
I'd suggest the .livequery() plugin for this still:
$(function() {
$('.cool').livequery(function() {
$(this).coolPlugin();
});
$('#cool_link').click(function(){
$('#cool_container').load('/anothercooldiv.html');
});
});
The important bit:
$('.cool').livequery(function() {
$(this).coolPlugin();
});
Will run for every current and future .cool element as they're added, running the plugin on each.
Applying the plugin to the newly ajax loaded content shouldn't be too tricky:
$('#cool_container').load('/anothercooldiv.html', function() {
$(this).coolPlugin();
});