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?

did any answer below help with your issue ? – N69S Feb 20 '20 at 08:53 — N69S, Feb 20 '20 at 08:53

score 424 · Answer 1 · edited Jul 11 '13 at 13:56

424

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

edited Jul 11 '13 at 13:56

Rubens Mariuzzo

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answered Jul 11 '13 at 11:14

Andreyco

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2

i tried this, failed :x – TransformBinary Jan 05 '14 at 07:58
5

In my case I was using a starting / for the URL, removing it would solve the issue – Joaquín L. Robles Jan 05 '15 at 18:17
@Andreyco what if I need to put a banner on all the pages except the home page try this not working `@if(Request::url()!=='/')
{{ HTML::image('images/fullimage3.jpg') }}
@endif` – Yousef Altaf Jun 11 '16 at 16:04
4

I use the next format for it:
, because the format with {{ some_code }} use string encoding.

Viktor

Oct 08 '16 at 20:55

3

`url()` [now returns a builder instance](https://laravel.com/docs/5.4/helpers#method-url). must use `url()->current()` – Jeff Puckett Mar 14 '17 at 18:45

@JeffPuckettII Could you provide Laravel version this change is related to? I will add note about this to answer. TY. – Andreyco Mar 15 '17 at 11:47

I'm not sure when it started, but the link I provided is for current 5.4 – Jeff Puckett Mar 15 '17 at 12:22

1

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. – Ryan Feb 19 '19 at 20:58

This leaves off the query string. Some answers below recommended `url()->full()`, which includes the query string. – Skeets Feb 06 '23 at 02:30

score 114 · Answer 2 · answered Jan 08 '15 at 17:36

114

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

answered Jan 08 '15 at 17:36

Naing Win

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50

FYI, `Route::is('testing')` is the same as `Route::current()->getName() == 'testing'`. – Hkan May 21 '15 at 17:31
@Hkan 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 – Naveen Singh raghuvanshi Apr 09 '16 at 10:01
1

@Naveen nope, `Route::is()` checks for the route name, not the path. – Hkan Apr 09 '16 at 10:35
1

also shorthand `Route::currentRouteName()` – Jeff Puckett Mar 14 '17 at 18:43
@Hkan Yeah, but that wouldn't work when working with parameters in your URL, for example, tokens ... – Pathros Aug 20 '18 at 17:52
@Pathros you can use wildcards like `Route::is('foo.*)` or even `Route::is('foo*')`. – Hkan Aug 20 '18 at 17:54

score 77 · Answer 3 · answered Jan 08 '15 at 14:27

77

Maybe you should try this:

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

answered Jan 08 '15 at 14:27

Dean Gite

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score 68 · Answer 4 · edited Feb 22 '19 at 20:14

68

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

edited Feb 22 '19 at 20:14

Zoe

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answered Apr 24 '17 at 13:51

Sagar Naliyapara

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4

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. – Spencer O'Reilly May 12 '17 at 16:23
``` @if(url()->current() == '/admin/search-result' ``` I used this code but it doesn't work? – Mahdi Safari Aug 15 '20 at 12:49
@MahdiSafari in that case just write `@if(Request::is('admin/search-result'))` – Sagar Naliyapara Aug 19 '20 at 03:29
This worked very nicely. Thanks – Ken Dec 07 '21 at 16:38

score 53 · Answer 5 · answered Jul 31 '14 at 21:30

53

I'd do it this way:

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

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

answered Jul 31 '14 at 21:30

Jucaman

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How to tell if the current dir is home? I mean, public/. Ahh, i got it, i just type `=='public'` – Pathros Feb 05 '15 at 18:52
1

If home dir is '/' in route.php, you just write `== '/')` – Jucaman Feb 17 '15 at 05:22
How would you make it work with the parameters as well? Not only the route name. – Pathros Feb 17 '17 at 02:01
You can concatenate parameters in many ways after route name, depending on how you configured your routes. – Jucaman Mar 22 '17 at 21:37
I do it like this: `Request::url()` - than you get the complete URL – Derk Jan Speelman Jan 15 '18 at 18:41

score 27 · Answer 6 · answered Jul 28 '16 at 04:31

27

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>

answered Jul 28 '16 at 04:31

Abduhafiz

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Thank you, I knew the code but Google got me here in 5 seconds haha. – Apr 12 '17 at 22:24
Also this same working with laravel 4.2 and it's working good. Thank you – Vipertecpro Feb 19 '18 at 06:49
Also working on Laravel 5.6, thanks... this just saved me from making navs for each view :D – NaturalDevCR Jun 03 '18 at 07:25

score 20 · Answer 7 · edited Feb 22 '19 at 20:14

20

A little old but this works in L5:

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

This captures both /mycategory and /mycategory/slug

edited Feb 22 '19 at 20:14

Zoe

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answered Mar 17 '16 at 16:06

VineFreeman

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I'm using `{{ Request::is('clientes/*') ? 'active' : ''}}` – Alex Angelico May 31 '20 at 21:16

score 18 · Answer 8 · answered May 05 '17 at 02:52

18

Laravel 5.4

Global functions

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

answered May 05 '17 at 02:52

Rob

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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 "/" – Ryan S Jul 16 '17 at 08:18
I used `request()->routeIs('...')` – Lachezar Todorov Jan 27 '20 at 14:56

score 17 · Answer 9 · edited Feb 22 '19 at 20:13

17

You can use this code to get current URL:

echo url()->current();

echo url()->full();

I get this from Laravel documents.

edited Feb 22 '19 at 20:13

Zoe

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answered Dec 16 '18 at 12:37

aydin abedinia

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score 17 · Answer 10 · answered Aug 05 '20 at 14:53

17

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

answered Aug 05 '20 at 14:53

Siva

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Alias · Answer 11 · 2013-07-11T11:22:47.517

15

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.

edited Jul 11 '13 at 11:22

answered Jul 11 '13 at 11:05

Alias

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score 14 · Answer 12 · answered Aug 01 '20 at 10:49

14

For me this works best:

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

answered Aug 01 '20 at 10:49

Slowwie

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score 12 · Answer 13 · answered Feb 15 '16 at 20:30

12

A simple navbar with bootstrap can be done as:

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

answered Feb 15 '16 at 20:30

Alejandro Silva

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score 10 · Answer 14 · answered Jun 23 '19 at 09:04

10

The simplest way is

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

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

answered Jun 23 '19 at 09:04

Naveen DA

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score 9 · Answer 15 · answered Jul 14 '15 at 11:38

9

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

But here is a complex way:

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

answered Jul 14 '15 at 11:38

shasi kanth

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score 9 · Answer 16 · answered Jan 07 '16 at 06:32

9

There are two ways to do that:

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

or

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

answered Jan 07 '16 at 06:32

gines.nt

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score 9 · Answer 17 · answered Jun 30 '18 at 10:18

9

You should try this:

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

answered Jun 30 '18 at 10:18

pardeep

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score 8 · Answer 18 · answered Feb 07 '19 at 19:42

8

For named routes, I use:

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

answered Feb 07 '19 at 19:42

Alex Pilyavskiy

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score 7 · Answer 19 · answered Nov 18 '16 at 17:42

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>

score 7 · Answer 20 · edited Feb 22 '19 at 20:12

7

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

edited Feb 22 '19 at 20:12

Zoe

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answered Feb 22 '19 at 20:11

mehran online

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Kenneth Sunday · Answer 21 · 2018-10-24T05:46:10.917

6

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.

edited Oct 24 '18 at 05:46

answered Nov 30 '17 at 14:14

Kenneth Sunday

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score 6 · Answer 22 · answered Oct 12 '19 at 09:15

6

In Blade file

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

answered Oct 12 '19 at 09:15

Faridul Khan

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score 4 · Answer 23 · answered Sep 05 '17 at 06:42

4

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

answered Sep 05 '17 at 06:42

Krishnamoorthy Acharya

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Mahdi Bashirpour · Answer 24 · 2022-12-14T02:47:15.000

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

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. — Joundill, Dec 14 '22 at 02:37

score 2 · Answer 25 · edited Feb 22 '19 at 20:13

2

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

edited Feb 22 '19 at 20:13

Zoe

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answered Sep 28 '18 at 17:13

C47

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Thank you for this code snippet, which might provide some limited, immediate help. A [proper explanation would greatly improve its long-term value](//meta.stackexchange.com/q/114762/206345) 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. – jhpratt Sep 28 '18 at 17:30

score 2 · Answer 26 · answered Mar 25 '21 at 16:51

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>

score 1 · Answer 27 · edited Feb 22 '18 at 07:07

1

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

edited Feb 22 '18 at 07:07

Pang

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answered Feb 22 '18 at 07:02

Alvin Davis Vadassery

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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. – rollstuhlfahrer Feb 22 '18 at 09:29

score 1 · Answer 28 · edited Feb 22 '19 at 20:13

1

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

 Request::url()

edited Feb 22 '19 at 20:13

Zoe

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answered Mar 07 '18 at 05:29

Ravindra Bhanderi

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score 1 · Answer 29 · answered Mar 25 '18 at 12:45

1

Try This:

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

answered Mar 25 '18 at 12:45

Rafiqul Islam

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score 0 · Answer 30 · edited Apr 12 '18 at 20:38

0

Try this way :

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

edited Apr 12 '18 at 20:38

msanford

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answered Apr 12 '18 at 12:25

parmod

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Add some explanation to your answer – Pankaj Makwana Apr 12 '18 at 14:10

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

30 Answers30

1. Check if URL = X

2. Check if URL contains X

3. Check route by its name

4. Check by routes names

Linked

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