Laravel - @yield() not allowed in if

Question

In my title I've been using @yield('title') for a while to create a dynamic title for every different page. I'd like to take it one step further and do something like this:

@if(@yield('title') == 'post') 
     <h1> This is the post page </h1>
@endif

But if I try that I get the error:

"The "yield" expression can only be used inside a function"

I've searched around and found things like Section::yield('title') but they all didn't work.

How can I use the value of @yield('title') in an if statement? I know you can send a $title to the view from the controller but I dont think that looks very good.. You would have to define the title twice that way.

Thanks for reading!

Answer 1

let's take a look at your code

@if(@yield('title') == 'post') 
 <h1> This is the post page </h1>
@endif

let's assume it works. then what does @section('title') will hold? Post? it doesn't make any sense.

@yield is used to replace the content of the section. it shouldn't be under conditional. it shouldn't be use as a variable. it has specific purpose. whatever conditionals you want to use, use it either in view or controller.

controller:

return View::make('view.blade.php',['title' => 'post']);

that's it. that's how you send a var to the view. i din't see anything wrong with the approach.

for laravel 4.1 and below, you could use this command.

@extends('layout.blade.php')->with('title','post')

but this behavior has been changed. till 4.1, the vars passed to the view wasn't accessible in layout. now, in 4.2, this behavior is changed and whatever data you pass to the view, it is also present in the controller.

Answer 2

What if you just define that as a section as well, it's cleaner that way. I'm assuming you have a blade layout that extends a master layout.

@extends('layouts.master')

@section('title', 'Some title')
@section('header')
     <h1> This is the post page </h1>
@stop

Then just yield them both in your master layout. If you have another page which should have a different title and header then you should create a new view that extends the master layout again.

I'm not sure of this is what you are trying to do but you can't have the content of the @yield as a condition to an @if. At least not without involving a PHP tag in there, which I think defeats the purpose of using a templating engine.

If it's not what you're looking for then at least I hope it helped.

See http://laravel.com/docs/4.2/templates for your reference.

Answer 3

Ok, I'm not saying this is pretty but you can pass PHP variables from the view to the layout (or parent view)

View containing "knowing" the title

<?php $title = 'post' ?>

Layout view

<?php
    // just making sure the variable is set
    if(!isset($title)){
        $title = 'default-title';
    }
?>

@if($title == 'post')
    <h1> This is the post page </h1>
@endif

However you may have to ask yourself if you should change the way you want to do this. (Maybe Integral Wind-up's answer helps). Also View::share('title', 'post') from your controller would be much cleaner.

Answer 4

like in this code example @if(View::getSection('title')=='Homepage') try using this.

@if(View::getSection('title')=='Anasayfa')

    <div class="hero_area">
        <!-- header section strats -->
        @include('home.header')
        <!-- end header section -->

        @include('home.slider')


    </div>
@else
    @include('home.header')
@endif