How to Get the Current URL Inside @if Statement (Blade) in Laravel 4?

Question

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?

Showiness

did any answer below help with your issue ? – Monotony

Answer 1

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

Inebriate

<li{{ (Request::is('admin/dashboard') ? ' class="active"' : '') }}> i tried this, failed :x – Ustulation

In my case I was using a starting / for the URL, removing it would solve the issue – Guv

@Inebriate what if I need to put a banner on all the pages except the home page try this not working @if(Request::url()!=='/') <div class="bannerImage">{{ HTML::image('images/fullimage3.jpg') }}</div> @endif – Grams

I use the next format for it: <li{!! (Request::url() == url('/')) ? ' class="active"' : '' !!}> , because the format with {{ some_code }} use string encoding. – Aerology

url() now returns a builder instance. must use url()->current() – Excoriate

@JeffPuckettII Could you provide Laravel version this change is related to? I will add note about this to answer. TY. – Inebriate

I'm not sure when it started, but the link I provided is for current 5.4 – Excoriate

Although I upvoted this a long time ago, today I was having trouble with it on Laravel 5.7 and eventually got this to work instead: @if (Request::path() != 'login'). Notice the lack of slash before "login", too. – Contraceptive

This leaves off the query string. Some answers below recommended url()->full(), which includes the query string. – Atrocious

Answer 2

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

Pornography

FYI, Route::is('testing') is the same as Route::current()->getName() == 'testing'. – Wellstacked

@Wellstacked No both are different, Route::is('testing') ->> testing it will not work Route::is('test') ->> test it will work Route::current()->getName() == 'testing' alias and url are differents – Fraternal

@Naveen nope, Route::is() checks for the route name, not the path. – Wellstacked

also shorthand Route::currentRouteName() – Excoriate

@Wellstacked Yeah, but that wouldn't work when working with parameters in your URL, for example, tokens ... – Epitome

@Epitome you can use wildcards like Route::is('foo.*) or even Route::is('foo*'). – Wellstacked

Answer 3

Maybe you should try this:

<li class="{{ Request::is('admin/dashboard') ? 'active' : '' }}">Dashboard</li>

Mongoloid

Answer 4

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

Unhopedfor

If you need to get the query string in the blade as well then you can use request()->getQueryString() which is very helpful in conjunction with url()->current() as that leaves off the query string. – Cichlid

``` @if(url()->current() == '/admin/search-result' ``` I used this code but it doesn't work? – Taille

@MahdiSafari in that case just write @if(Request::is('admin/search-result')) – Unhopedfor

This worked very nicely. Thanks – Streptococcus

Answer 5

I'd do it this way:

@if (Request::path() == '/view')
    // code
@endif

where '/view' is view name in routes.php.

Airlie

How to tell if the current dir is home? I mean, public/. Ahh, i got it, i just type =='public' – Epitome

If home dir is '/' in route.php, you just write == '/') – Airlie

How would you make it work with the parameters as well? Not only the route name. – Epitome

You can concatenate parameters in many ways after route name, depending on how you configured your routes. – Airlie

I do it like this: Request::url() - than you get the complete URL – Whoremaster

Answer 6

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>

Unweighed

Thank you, I knew the code but Google got me here in 5 seconds haha. – Hong

Also this same working with laravel 4.2 and it's working good. Thank you – Galloon

Also working on Laravel 5.6, thanks... this just saved me from making navs for each view :D – Ta

Answer 7

A little old but this works in L5:

<li class="{{ Request::is('mycategory/', '*') ? 'active' : ''}}">

This captures both /mycategory and /mycategory/slug

Willette

I'm using {{ Request::is('clientes/*') ? 'active' : ''}} – Leandroleaning

Answer 8

Laravel 5.4

Global functions

@if (request()->is('/'))
    <p>Is homepage</p>
@endif

Avowed

This does not always work if you're dealing with query string like so domain.com/?page_id=1, domain.com/?page_id=2, as those URLs also equal as "/" – Untoward

I used request()->routeIs('...') – Gustav

Answer 9

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

Culminant

Answer 10

You can use this code to get current URL:

echo url()->current();

echo url()->full();

I get this from Laravel documents.

Seismoscope

Answer 11

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.

Nesline

Answer 12

For me this works best:

class="{{url()->current() == route('dashboard') ? 'bg-gray-900 text-white' : 'text-gray-300'}}"

Whaling

Answer 13

A simple navbar with bootstrap can be done as:

    <li class="{{ Request::is('user/profile')? 'active': '' }}">
        <a href="{{ url('user/profile') }}">Profile </a>
    </li>

Pape

Answer 14

The simplest way is

<li class="{{ Request::is('contacts/*') ? 'active' : '' }}">Dashboard</li>

This colud capture the contacts/, contacts/create, contacts/edit...

Moonseed

Answer 15

The simplest way is to use: Request::url();

But here is a complex way:

URL::to('/').'/'.Route::getCurrentRoute()->getPath();

Genuflect

Answer 16

There are two ways to do that:

<li{!!(Request::is('your_url')) ? ' class="active"' : '' !!}>

or

<li @if(Request::is('your_url'))class="active"@endif>

Blotch

Answer 17

You should try this:

<b class="{{ Request::is('admin/login') ? 'active' : '' }}">Login Account Details</b>

Vina

Answer 18

For named routes, I use:

@if(url()->current() == route('routeName')) class="current" @endif

Mckee

Answer 19

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>

Chorion

Answer 20

class="nav-link {{ \Route::current()->getName() == 'panel' ? 'active' : ''}}"

Argue

Answer 21

In Blade file

@if (Request::is('companies'))
   Companies name 
@endif

Cota

Answer 22

1. Check if URL = X

Simply - you need to check if URL is exactly like X and then you show something. In Controller:

if (request()->is('companies')) {
  // show companies menu or something
}

In Blade file - almost identical:

@if (request()->is('companies'))
  Companies menu
@endif

2. Check if URL contains X

A little more complicated example - method Request::is() allows a pattern parameter, like this:

if (request()->is('companies/*')) {
  // will match URL /companies/999 or /companies/create
}

3. Check route by its name

As you probably know, every route can be assigned to a name, in routes/web.php file it looks something like this:

Route::get('/companies', function () {
  return view('companies');
})->name('comp');

So how can you check if current route is 'comp'? Relatively easy:

if (\Route::current()->getName() == 'comp') {
  // We are on a correct route!
}

4. Check by routes names

If you are using routes by names, you can check if request matches routes name.

if (request()->routeIs('companies.*')) {
  // will match routes which name starts with companies.
}

Or

request()->route()->named('profile')

Will match route named profile. So these are four ways to check current URL or route.

source

Ankledeep

The question is about Laravel 4, and this is based on a tutorial for Laravel 9. Request::is('companies') is how it's done in Laravel 4. – Roband

Answer 23

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.

Chalcography

Answer 24

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

Snakebird

Answer 25

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

Avivah

Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made. – Hackery

Answer 26

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>

Tinsel

Answer 27

@if(request()->path()=='/path/another_path/*')
@endif

Hindward

While this code may answer the question, providing additional context regarding why and/or how this code answers the question improves its long-term value. – Alejandraalejandrina

Answer 28

There are many way to achieve, one from them I use always

 Request::url()

Foothill

Answer 29

Try This:

<li class="{{ Request::is('Dashboard') ? 'active' : '' }}">
    <a href="{{ url('/Dashboard') }}">
	<i class="fa fa-dashboard"></i> <span>Dashboard</span>
    </a>
</li>

Supraliminal

Answer 30

Try this way :

<a href="{{ URL::to('/registration') }}">registration </a>

Kr

Add some explanation to your answer – Beulahbeuthel