How to Get the Current URL Inside @if Statement (Blade) in Laravel 4?
LaravelLaravel 4Laravel BladeLaravel Problem Overview
I am using Laravel 4. I would like to access the current URL inside an @if condition in a view using the Laravel's Blade templating engine but I don't know how to do it.
I know that it can be done using something like <?php echo URL::current(); ?> but It's not possible inside an @if blade statement.
Any suggestions?
Laravel Solutions
Solution 1 - Laravel
You can use: Request::url() to obtain the current URL, here is an example:
@if(Request::url() === 'your url here')
// code
@endif
Laravel offers a method to find out, whether the URL matches a pattern or not
if (Request::is('admin/*'))
{
// code
}
Check the related documentation to obtain different request information: http://laravel.com/docs/requests#request-information
Solution 2 - Laravel
You can also use Route::current()->getName() to check your route name.
Example: routes.php
Route::get('test', ['as'=>'testing', function() {
return View::make('test');
}]);
View:
@if(Route::current()->getName() == 'testing')
Hello This is testing
@endif
Solution 3 - Laravel
Maybe you should try this:
<li class="{{ Request::is('admin/dashboard') ? 'active' : '' }}">Dashboard</li>
Solution 4 - Laravel
To get current url in blade view you can use following,
<a href="{{url()->current()}}">Current Url</a>
So as you can compare using following code,
@if (url()->current() == 'you url')
//stuff you want to perform
@endif
Solution 5 - Laravel
I'd do it this way:
@if (Request::path() == '/view')
// code
@endif
where '/view' is view name in routes.php.
Solution 6 - Laravel
This is helped to me for bootstrap active nav class in Laravel 5.2:
<li class="{{ Request::path() == '/' ? 'active' : '' }}"><a href="/">Home</a></li>
<li class="{{ Request::path() == 'about' ? 'active' : '' }}"><a href="/about">About</a></li>
Solution 7 - Laravel
A little old but this works in L5:
<li class="{{ Request::is('mycategory/', '*') ? 'active' : ''}}">
This captures both /mycategory and /mycategory/slug
Solution 8 - Laravel
Laravel 5.4
Global functions
@if (request()->is('/'))
<p>Is homepage</p>
@endif
Solution 9 - Laravel
I personally wouldn't try grabbing it inside of the view. I'm not amazing at Laravel, but I would imagine you'd need to send your route to a controller, and then within the controller, pass the variable (via an array) into your view, using something like $url = Request::url();.
One way of doing it anyway.
EDIT: Actually look at the method above, probably a better way.
Solution 10 - Laravel
You can use this code to get current URL:
echo url()->current();
echo url()->full();
I get this from Laravel documents.
Solution 11 - Laravel
A simple navbar with bootstrap can be done as:
<li class="{{ Request::is('user/profile')? 'active': '' }}">
<a href="{{ url('user/profile') }}">Profile </a>
</li>
Solution 12 - Laravel
For me this works best:
class="{{url()->current() == route('dashboard') ? 'bg-gray-900 text-white' : 'text-gray-300'}}"
Solution 13 - Laravel
You will get the url by using the below code.
For Example your URL like https//www.example.com/testurl?test
echo url()->current();
Result : https//www.example.com/testurl
echo url()->full();
Result: https//www.example.com/testurl?test
Solution 14 - Laravel
The simplest way is to use: Request::url();
But here is a complex way:
URL::to('/').'/'.Route::getCurrentRoute()->getPath();
Solution 15 - Laravel
There are two ways to do that:
<li{!!(Request::is('your_url')) ? ' class="active"' : '' !!}>
or
<li @if(Request::is('your_url'))class="active"@endif>
Solution 16 - Laravel
You should try this:
<b class="{{ Request::is('admin/login') ? 'active' : '' }}">Login Account Details</b>
Solution 17 - Laravel
The simplest way is
<li class="{{ Request::is('contacts/*') ? 'active' : '' }}">Dashboard</li>
This colud capture the contacts/, contacts/create, contacts/edit...
Solution 18 - Laravel
Set this code to applied automatically for each <li> + you need to using HTMLBuilder library in your Laravel project
<script type="text/javascript">
$(document).ready(function(){
$('.list-group a[href="/{{Request::path()}}"]').addClass('active');
});
</script>
Solution 19 - Laravel
For named routes, I use:
@if(url()->current() == route('routeName')) class="current" @endif
Solution 20 - Laravel
In Blade file
@if (Request::is('companies'))
Companies name
@endif
Solution 21 - Laravel
Another way to write if and else in Laravel using path
<p class="@if(Request::is('path/anotherPath/*')) className @else anotherClassName @endif" >
</p>
Hope it helps
Solution 22 - Laravel
instead of using the URL::path() to check your current path location, you may want to consider the Route::currentRouteName() so just in case you update your path, you don't need to explore all your pages to update the path name again.
Solution 23 - Laravel
class="nav-link {{ \Route::current()->getName() == 'panel' ? 'active' : ''}}"
Solution 24 - Laravel
@if(request()->path()=='/path/another_path/*')
@endif
Solution 25 - Laravel
Try this:
@if(collect(explode('/',\Illuminate\Http\Request::capture()->url()))->last() === 'yourURL')
<li class="pull-right"><a class="intermitente"><i class="glyphicon glyphicon-alert"></i></a></li>
@endif
Solution 26 - Laravel
There are many way to achieve, one from them I use always
Request::url()
Solution 27 - Laravel
Try This:
<li class="{{ Request::is('Dashboard') ? 'active' : '' }}">
<a href="{{ url('/Dashboard') }}">
<i class="fa fa-dashboard"></i> <span>Dashboard</span>
</a>
</li>
Solution 28 - Laravel
For Laravel 5.5 +
<a class="{{ Request::segment(1) == 'activities' ? 'is-active' : ''}}" href="#">
<span class="icon">
<i class="fas fa-list-ol"></i>
</span>
Activities
</a>
Solution 29 - Laravel
Try this way :
<a href="{{ URL::to('/registration') }}">registration </a>