I can not find a solution how to include blade template using button click.
For example lets say I have a main view:
<div id="HEREGOESMYTEMPLATES">
</div>
<div class="row">
<button onclick="addTemplate()">ADD TEMPLATE</button>
</div>
Then script, which should include template
<script type="app/js">
function addTemplate()
{
//Get template and include into div with id HEREGOESMYTEMPLATE
}
</script>
And then the template itself (lets say partial.blade.php):
<div>Include me to main blade file</div>
Is there any possible way by including it without creating every tag in JS and then appending to the div?
Thank you.
UPDATE.
As blade template is not supposed to do such work as it is rendered in backend, I used VueJS to this, which I believe is the best solution in this situation.
I have a layout contain header include navbar and slider image.
My views folder has structure like:
+ layouts
- index.blade.php
- master.blade.php
+ posts
- list.blade.php
+ partials
- navbar.blade.php
- footer.blade.php
Note:
+ is a folder.
- is blade template file
Normally, in the master.blade.php is declare css/js file, include navbar and footer blade template.
But navbar inside a div#home, this element contains navbar and slider.
<section id="home">
#include('partials.navbar')
<!-- slider start -->
#yield('slider_images')
<!-- slider end -->
</section>
So, I can't split navbar to insert into master.blade.php. You can see master.blade.php like:
//add some js and css.
<body>
#yield('content')
#include('partials.footer')
</body>
I don't add line: #include('partials.navbar') because it include a slide.
The temporary solution is to add it into index.blade.php.
But the problem is: I have a child page, it #extends('layouts.master'). But the layout.master don't have contained partials.navbar. It only needs navbar and footer, don't need the slider.
Have any method to do this without change structure of HTML file.
You could pas a variable to the blade template:
#include('partials.navbar', ['slider' => true])
and then in the navbar you could check for that variable
#if(isset($slider) && $slider === true)
<!-- slider start -->
#yield('slider_images')
<!-- slider end -->
#endif
in that way the slider will be shown only when the slider variable is set to true
I'm trying to yield a section inside another section. But this does not work as expected, I see blank output.
#section('3show')
This is a text string
#stop
#section('page-content')
<div id="content">
<article>
#yield('3show')
</article>
</div>
<!--#content-->
#stop
Any ideas to yield section inside another section ?
Ok, this is what I tried and I can confirm that this works, at least for Laravel 5+ (I have L5.2). This is how I suggest you to use your blade templates.
Lets start saying that to yield a section into another section you have to define your included section before container section definition. So, with that clear, I solved this situation like this:
I got a main blade (main.blade.php) template which has something like:
<section class="content">
<!-- Your Page Content Here -->
#yield('main-content')
</section><!-- /.content -->
I got a second blade (common.blade.php) template which has that common stuff you may want to show across many pages and where main-content section is defined. This one looks like:
#section('main-content')
<div class="container">
#yield('extra-content')
</div>
#endsection
Finally I got a 3rd template (test.blade.php) which extend the main template and include the common stuff I want to show, but be careful because the order matters. This one looks like:
#extends('main')
#section('extra-content')
<div>
<span> This is a test! </span>
</div>
#endsection
#include('common')
In your controller or your route (wherever you return your view), you should return the 3rd template.
In my projects i create some partials in order to have cleaner code and i give them as an example a name : 3show.blade.php. In order to use them in a section i just include them.
I think this will do what you want.
#section('content')
#include('3show.blade.php')
#endsection
I had the same issue.
You can't use the #extends option in this case, you need to use #include .
So lets say you have:
the root layout (layouts/app.blade.php)
an extra layout (layouts/extra.blade.php)
the actual view that you are calling (someview.blade.php)
The solution is to use add/inherit the root layout (1) to the top line of your view (3):
#extends('layouts.app')
Next, add/inherit the extra layout (2) to the second line of your view, BUT use #include not #extends:
#include('layouts.extra')
...and remember to wrap the content of your extra layout in an #section name, for example #section('extra')
Finally you can call your extra layout content from your view:
<p>Look below, you will see my extra content...</p>
#yield('extra')
So in summary, your someview.blade.php view would look like:
#extends('layouts.app')
#include('layouts.extra')
#section('content')
<p>Look below, you will see my extra content...</p>
#yield('extra')
#endsection
solution 1:
you can use #show instead of #stop at the end of section
then laravel will render your #yield parts ...
like this :
#section('page-content')
<div id="content">
<article>
#yield('3show')
</article>
</div>
#show # not #stop
#section('3show')
This is a text string
#stop
this way render you view and show result
so if you cll your section for twice then result will be shoed twice
solution 2:
insert call section before yiel it
like this :
**first call section :**
#section('3show')
This is a text string
#stop
**then define your section : 3show**
#section('page-content')
<div id="content">
<article>
#yield('3show')
</article>
</div>
#stop **not #show**
assume you have two files:
-article_base.blade.php -> the default data in every article.
-article_index.blade.php -> the customized file.
article_base.blade.php
#extends('layout')
#section('page-content')
<div id="content">
<article>
....
#yield('3show')
....
</article>
</div>
<!--#content-->
#stop
article_index.blade.php
#extends('article_base')
#section('3show')
This is a text string
#endsection
I hope this works
You have to both #include and #yield the child template in your main template to be able to add the child template at a specific place inside the main template (not just before or after the main template - this is done by adding #parent in the child template - but in between):
main.blade.php
#include('child')
<div>
...
#yield('child')
</div>
child.blade.php
#section('child')
<div>
...
</div>
#endsection
How can we use function anchor of codeigniter for div tag?
I got div around which I want to put anchor(uri segments, text, attributes)
in pure html it looks like this:
<a class="a_services" href="http://justinbieber.com">
<div id="seminar">Hello world</div>
</a>
You can include code in the second parameter.
echo anchor('http://justinbieber.com', '<div id="seminar">Hello world</div>', 'class="a_services"');
Let's say that I have two html pages that are identically designed, but have different content. I have the same div with the same id on both pages. How do I use jQuery.load (or what do I use) so that the div#conent does not get added into the div#content of the first page.
I've tried this:
$(document).ready(function(){
$("a#linkHome").click(function(){$("div#content").load('index.htm #content');});
$("a#linkPage2").click(function(){$("div#content").load('page2.htm #content');});
});
... but it ends up adding another div to the already existing div!
<div id="content">
<div id="content">
Blah Blah Blah
<div id="content">
</div>
Try with:
$(document).ready(function(){
$("a#linkHome").click(function(){$("div#content").load('index.htm #content *');});
$("a#linkPage2").click(function(){$("div#content").load('page2.htm #content *');});
});
in this way you get all elements inside the div#content but not the div itself.
Or you can try the opposite approach. Just add a wrapper div into your target page.