blade

问题 I am looking to Bootstrap my site and as such, have put common twitter bootstrap components into blade templates. sidebar.blade.php @include('panel1') @include('panel2') panelTemplate.blade.php <div class="panel panel-primary"> <div class="panel-heading"> <div class="panel-title"> @yield('title') </div> </div> <div class="panel-body"> @yield('body') </div> <div class="panel-footer"> @yield('footer') </div> </div> This way, every time I wish to use a panel, then I can use @extends(

问题 I had a user submit a form, and some of the fields were dropdowns like so <div class="form-group row"> <label for="type" class="col-md-3 col-form-label text-md-right">Type</label> <div class="col-md-9"> <select class ="form-control" id="type" name="type"> <option>Apartment</option> <option>House</option> <option>Studio</option> <option>Flat</option> </select> @if ($errors->has('type')) <span class="invalid-feedback"> <strong>{{ $errors->first('type') }}</strong> </span> @endif </div> </div>

问题 My problem is that I need to pass string as an argument to view, which contains the Blade Template tags to be rendered. I've done my research and I believe the other problems asked don't relate to mine. Hence following is what I want to achieve. I want to retrieve the master layout from database and extend the child views from it. So in my case, the view that is to be passed to view() method needs to be extended from a string in database. So something like the following should work in page

问题 TL;DR: What is the correct way to link to a stylesheet in Laravel 5? Background: I'm using the dev version of Laravel 4.3 (5) because I want to use Socialite, and it makes sense to develop with that from the start. I am having a problem getting templates transferred from 4.2 I've moved my blade layout files to the new directory structure (resources/templates) and placed my CSS in the public/css/ folder. When I load my route /test/ all I get is "Whoops, looks like something went wrong." For

问题 I have Completely assigned the Passport REST API and i get token and tested on PostMan and it retrieve the data with this Auth this is normal js html with laravel blade view (NOT VUE.JS) <script> $.ajax({ method: 'POST', dataType: 'json', url: "**********/api/User/GetPost", headers: { 'Authorization':'Bearer Tokenblablabla', 'Content-Type':'application/json' }, data: null, success: function(data){ console.log('succes: '+data); } }); this is the controller function public function GetPosts

问题 Suppose this is a JS code and I want to print the value of id . Here id is a variable with a value receiving into the jq function. function follow_or_unfollow(id,action){ myUrl = "{{ action('FollowsController@show', 'id') }}" ; } If we mension 'id' it will show the id string. PS. here {{ is using as it meant as php code in blade template. Now i need to print the script variable inside a php code. 回答1: I am afraid that what you are asking is not possible. The simple reason being JavaScript is

问题 I have a for loop that I want to include a template view in: table.blade: #game h1= table.name -if( table.playerSitting() ) a.btn.btn-danger.stand(href="#") Stand #table -for (var i=0; i<7;i++) include 'seat' exposing i seat.blade: - id = 'p' + (i+1) div(id="#{id}") -if (table.hasAt(i)) p= table.usernameAt(i) -else -if( !table.playerSitting() ) a.btn.btn-inverse.sit(href="#", data-key="#{i}") Sit -else p   When I first load the view, everything draws fine. But if I sit or stand, or go to

问题 In my Laravel project I'm trying to load an image into my view using blade but it displays as a broken link. I inspect it with firebug and the src is pointing to the image but nothing is displaying. My image is located in my project's public/images/users/3/profilePictures/ Here is my <img> in the view.blade.php <img class="com-profile-picture" src="images/users/{{ $follower->id }}/profilePictures/50thumb-{{ $follower->profilePicture->url }}" alt="{{ $follower->first_name }} {{ $follower->last

问题 I have this jquery $(".waitingTime .button").click(function () { alert("Ema"); }); I have a a tag like this: <a href="{{ URL::to('restaurants/20') }}"></a> Can I do the same href action in the jquery function? Many Thanks 回答1: yes this is possible and has nothing to do with Laravel. There are different possibilities. If your query is embedded within the same laravel view, you put the URL directly in your jQuery code, for example like this: $(".waitingTime .button").click(function () { window

问题 I am trying to fetch value of the password field and display it in current form for one of the update pages {{ Form::label('Password')}} {{ Form::password('password',array('class' => 'form-control') }} Laravel blade's password field syntax does not allow parameter for input like it does for text fields for example, {{ Form::label('Email or Username')}} {{ Form::text('email',$useremaildata,array('class' => 'form-control')) }} Now I found one solution of using placeholders like that but it is