I've a master layout like this:
<head>
#yield('styles')
</head>
<body>
#include('header')
<div class="container-fluid">
#yield('content')
</div>
#yield('scripts')
</body>
Now I've following structure in page content:
#section('content')
<div class="page-content">
#include('sidebarandfooter')
</div>
#endsection
#section('copyright')
#include('copyrightv2')
#endsection
Sidebarandfooter.blade.php have following:
...[CODE for sidebar]...
#yield('copyright')
It's should be called inside <div class="container-fluid"></div> as I've different class of div container for different pages.
I'm not able to yield the copyright part. I've different copyright section for different pages. Is it wrong, how can we execute such kind?
Unfortunattely it's imposible in way you declare it because this section:
#section('copyright')
#include('copyrightv2')
#endsection
been loaded before you call this: #include('sidebarandfooter') with copyright section inside.
What you can do is to pass a key as a parameter to view to the included page content partial like this:
#section('content')
<div class="page-content">
#include('sidebarandfooter', ['copyrightsView' => 'copyrightv2'])
</div>
#endsection
and then just call inside Sidebarandfooter.blade.php in the include:
...[CODE for sidebar]...
#include($copyrightsView)
change your master layout like
<head>
#yield('styles')
</head>
<body>
#include('header')
<div class="container-fluid">
#yield('content')
</div>
#yield('copyright')
#yield('scripts')
</body>
since i am not seeing a section for copyright in your master layout
Related
I have a newly started project in Laravel 8.0, and I have a problem with views.
I have the following master.blade.
<html>
<head>
<title>App Name - #yield('title')</title>
</head>
<body>
<div class="container">
#yield('content')
</div>
<footer class="row">
#include('layouts.footer')
</footer>
</body>
</html>
I then have a view that extends that master blade:
#extends('layouts.app')
#extends('layouts.master')
#section('content')
HELLO
#stop
The problem is, when rendering the view, the information is duplicated. Appears once inside the container div so the #yield works, but is re-rendered in another div outside the container (main class="py-4").
Let's see if someone can help me with the problem.
Thank you very much in advance.
#extends('layouts.master')
#section('content')
HELLO
#endsection
you extends two time
You must add one extends, edit the code and do the code below
#extends('layouts.master')
#section('content','your title')
HELLO
#endsection
that the right way to get the view.
Im trying to render a webapge and use the thymleaf attribute "url" added with model.addAttribute but the Attribute is not beeing displayed on the html document.
My document.html file path is here:
/templates/webpage/document.html
#RequestMapping(value = "/webpage/document")
public String document(HttpServletRequest req, Model model) {
model.addAttribute("dialogurl", url);
return "/webpage/document";
}
Here is the html document
<!DOCTYPE html>
<html xmlns:th="http://www.thymeleaf.org" th:include="wrapperdialog :: page">
<head>
<title></title>
</head>
<body>
<div th:fragment="content">
<div class="container dialogpage">
<div class="row">
<div class="col-md-12">
<div id="typeform" th:attr="data-url=*{dialogurl}">
</div>
</div>
</div>
</div>
</div>
Please use this expression to bind dialog url: #{${dialogurl}}
<div id="typeform" th:attr="data-url=#{${dialogurl}}">
# prefix is used to specify a link and $ prefix is used to bind your model value.
Use $ to bind data.
<div id="typeform" th:attr="data-url=${dialogurl}">
Im with some doubts about how to strucure the views folder on laravel. For example a blog site where we have this structure in the homepage:
Header (where we have a menu with: a link to homepage, search bar to search for posts, sign in and login buttons)
Then a section with the last 5 posts.
Then a section with the most viewed posts.
And then a footer.
How we can structure this homepage in terms of views? For example maybe we can something like this:
-views
- layouts
- header.blade.php
- footer.blade.php
- posts
- lastposts.blade.php
- mostviewed.blade.php
- single.blade.php
- layout.blade.php
Do you think its ok? Because Im testing this and it is not working. For example in the layout.blade.php I have:
<!DOCTYPE html>
<html lang="en">
<head>
....
<link href="css/app.css" rel="stylesheet">
</head>
<body>
#include('layouts.nav')
#yield('lastposts')
#yield('mostviewed')
#include(layouts.footer')
</body>
</html>
And I get the header and the footer when I access the page corretly, but the both #yields dont show nothing.
In the lastposts.blade.php to show the last 5 posts I have:
#extends('layout')
#section('lastposts')
<div>
<h4>Title</h4>
<p>Text</p>
</div>
</div>
<div>
<h4>Title</h4>
<p>Text</p>
</div>
</div>
<div>
<h4>Title</h4>
<p>Text</p>
</div>
</div>
<div>
<h4>Title</h4>
<p>Text</p>
</div>
</div>
<div>
<h4>Title</h4>
<p>Text</p>
</div>
</div>
#endsection
I have the same logic for the mostviewed.blade.php, but again, the #yield('mostviewed') dont show nothing.
Do you know what is not corret?
Routes file:
Route::get('/', function () {
return view('layout');
}
I have just this route because I just have the homepage for now.
You created the wrong view. The parent view is layout, it doesn't know about mostviewed and lastposts. That's why you wrote #extends('layout'), that way when you create your content view, it will know that it has to extend stuff from layout.
Route::get('/', function () {
return view('posts.mostviewed');
}
I have the following files:
foo.blade.php
<html>
<head>
</head>
<body>
<h1>Foo Template</h1>
#yield('content')
</body>
</html>
bar.blade.php
<h2>Bar Template</h2>
<div class="bar-content">
#yield('bar-content')
</div>
I want to create another file that is able to extend both the above templates. e.g something like this:
#extends('foo')
#section('content')
<p>Hello World</p>
#extends('bar')
#section('bar-content')
<p>This is in div.bar-content</p>
#endsection
#endsection
To give:
<html>
<head>
</head>
<body>
<h1>Foo Template</h1>
<p>Hello World</p>
<h2>Bar Template</h2>
<div class="bar-content">
<p>This is in div.bar-content</p>
</div>
</body>
</html>
How can I do this?
Use multiple files.
for eg;
layout.blade.php:
#include('header')
#yield('layout_content')
#include('footer')
second.blade.php
#extends('layout')
#section('layout_content')
<div>
#yield('second_content')
</div>
#stop
third.blade.php
#extends('second.blade.php')
#section('second_content')
<h1>Hello World!</h1>
#stop
You can include #yield in any of the parent files and can be used in the child
I would recommend using components. Here is an example.
It seems to me layout/include has a weird logic when there are many of them and you start nesting them. Components are pretty straight forward. For these nested structures you are building there, components also have slots.
I think it's not possible to do what you want here, at least without extending Blade:
https://laravel.com/docs/5.1/blade#extending-blade
If I were you, I'd rearchitectured my views hierarchy to keep things simple.
I do not know if this will work, but try #include instead of #extends for your bar template. The put your section for bar below the other section (not nested). I did not tested this, so I hope it works ;)
// EDIT:
Try it with an if-statement in your foo file:
<html>
<head>
</head>
<body>
<h1>Foo Template</h1>
#yield('content')
#if(isset($displayBar) && $displayBar == true)
#include('dashboard.test.bar')
#endif
</body>
</html>
And now the child view:
#extends('dashboard.test.foo')
#section('content')
<p>Hello World</p>
#endsection
<?php $displayBar = true ?>
#section('bar-content')
<p>This is in div.bar-content</p>
#endsection
#php
$ext = '';
#endphp
#if (Request::segment(1)=='explications')
#php
$ext = '_explications'
#endphp
#endif
#extends('pages_templates/template_page_fluid'.$ext)
I can't looking for solution or understanding solution when I try use blade (laravel 4.2) for use with several sections.
For normal use (home extendes layout) not problem.
views/home.blade.php
#extends('layout')
#section('content')
Show content of home page
#endsection
view/layout.blade.php
<html>
<!-- Code html such head, template header... -->
...
#yield('content')
<!-- Rest code html such footer -->
...
</body>
</html>
But when I like move Rest code (footer) to another template file footer.blade.php I don't understand and try several ways.
view/footer.blade.php
#extends('layout')
#section('footer')
<hr>
<footer>
<p>© Tamainout Hébergement SARL 2015</p>
</footer>
</div>
#endsection
and change views/home.blade.php
<html>
<!-- Code html such head, template header... -->
...
#yield('content')
#yield('footer')
<!-- Rest code html such footer -->
...
</body>
</html>
But file footer.blade.php it isn't processed by laravel.
Apreciate some help
NOTE: I put code on github, abkrim/blade (only files envolved)
I think you have a syntax error in the views/home.blade.php.
Try:
#yield('footer')
instead of
#yield(footer')
EDIT:
include footer.blade.php in the layout.blade.php and remove extends from footer.blade.php
views/home.blade.php:
#extends('layout')
#section('content')
Show content of home page
#endsection
views/footer.blade.php:
<hr>
<footer>
<p>© Tamainout Hébergement SARL 2015</p>
</footer>
</div>
views/layout.blade.php:
#yield('content')
#include('footer')