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?
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
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
Route::is('testing')
is the same as Route::current()->getName() == 'testing'
.
Route::is('testing')
->> testing it will not work Route::is('test')
->> test it will work Route::current()->getName() == 'testing'
alias and url are differents
Route::is()
checks for the route name, not the path.
Route::currentRouteName()
Maybe you should try this:
<li class="{{ Request::is('admin/dashboard') ? 'active' : '' }}">Dashboard</li>
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
request()->getQueryString()
which is very helpful in conjunction with url()->current()
as that leaves off the query string.
@if(Request::is('admin/search-result'))
I'd do it this way:
@if (Request::path() == '/view')
// code
@endif
where '/view' is view name in routes.php.
=='public'
== '/')
Request::url()
- than you get the complete URL
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>
A little old but this works in L5:
<li class="{{ Request::is('mycategory/', '*') ? 'active' : ''}}">
This captures both /mycategory and /mycategory/slug
{{ Request::is('clientes/*') ? 'active' : ''}}
Laravel 5.4
Global functions
@if (request()->is('/'))
<p>Is homepage</p>
@endif
request()->routeIs('...')
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.
You can use this code to get current URL:
echo url()->current();
echo url()->full();
I get this from Laravel documents.
A simple navbar with bootstrap can be done as:
<li class="{{ Request::is('user/profile')? 'active': '' }}">
<a href="{{ url('user/profile') }}">Profile </a>
</li>
For me this works best:
class="{{url()->current() == route('dashboard') ? 'bg-gray-900 text-white' : 'text-gray-300'}}"
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
The simplest way is to use: Request::url();
But here is a complex way:
URL::to('/').'/'.Route::getCurrentRoute()->getPath();
There are two ways to do that:
<li{!!(Request::is('your_url')) ? ' class="active"' : '' !!}>
or
<li @if(Request::is('your_url'))class="active"@endif>
You should try this:
<b class="{{ Request::is('admin/login') ? 'active' : '' }}">Login Account Details</b>
The simplest way is
<li class="{{ Request::is('contacts/*') ? 'active' : '' }}">Dashboard</li>
This colud capture the contacts/, contacts/create, contacts/edit...
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>
For named routes, I use:
@if(url()->current() == route('routeName')) class="current" @endif
In Blade file
@if (Request::is('companies'))
Companies name
@endif
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
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.
class="nav-link {{ \Route::current()->getName() == 'panel' ? 'active' : ''}}"
@if(request()->path()=='/path/another_path/*')
@endif
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
There are many way to achieve, one from them I use always
Request::url()
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>
Try this way :
<a href="{{ URL::to('/registration') }}">registration </a>
Success story sharing
url()
now returns a builder instance. must useurl()->current()
@if (Request::path() != 'login')
. Notice the lack of slash before "login", too.