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posted 10 years ago
Views Blade
Last updated 2 years ago.
0

Hi,

as far as i know you can't use html tags with link_to_route function. So you have two options. First, you can use your first code with route($route) - it generates url for given route -. Second, you can create a new helper method something like link_to_route_icon() and use it :)

Lastly, you can learn how to write with mark up in here : http://laravel.io/forum/01-31-2014-how-to-mark-up-forum-posts

Last updated 2 years ago.
0

Thanks for the markdown link. I did look for that prior to posting but didn't find it. ;)

I figured as much about the html tags, as I get the same text display instead of rendered HTML using the HTML Builder. However I just thought it was pretty odd that the source shows the same code yet only one is rendered correctly.

<li class="has-form">
    <a class="small button" href="http://morris.local:8000/logout">
        <i class="fa fa-sign-out fa-lg"></i> Log Out
    </a>
</li>
<li class="has-form">
    <a class="small button" href="http://morris.local:8000/logout">
        <i class="fa fa-sign-out fa-lg"></i>

             Log Out
                            

    </a>
</li>

I already have a helpers.php file autoloaded using PSR-4, I'll probably go that route. Thanks for the info!

Last updated 2 years ago.
0

This should work

{{ HTML::decode(link_to_route('logout', '<i class="fa fa-sign-out"></i> Log Out', array(), ['class' => 'small button'])) }}
Last updated 9 years ago.
0

Thanks TorchSK Its work like a boss :D

TorchSK said:

This should work

{{ HTML::decode(link_to_route('logout', '<i class="fa fa-sign-out"></i> Log Out', array(), ['class' => 'small button'])) }}
0

this work, too.. {!! HTML::decode(link_to_route('logout', ''<i class="fa fa-sign-out"></i> Log Out', array(), ['class' => 'small button'])) !!}

Last updated 8 years ago.
0

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