如何在laravel 5中获取当前的路由名称?

I am trying to get current route name using Laravel 5.

我正在尝试用Laravel 5获得当前的路线名称。

In laravel 4 i was able to do it using Route::currentRouteName().

在laravel 4中，我可以使用Route::currentRouteName()来完成。

How can i do it in laravel 5?

我怎么用laravel 5呢?

Thanks

谢谢

15 个解决方案

#1

252

Try this

试试这个

Route::getCurrentRoute()->getPath();

或

\Request::route()->getName()

from v5.+

从v5 +。

use Illuminate\Support\Facades\Route;
$currentPath= Route::getFacadeRoot()->current()->uri();

Laravel 5.3

Laravel 5.3

Route::currentRouteName(); //use Illuminate\Support\Facades\Route;

Or if you need the action name

或者如果你需要动作名。

Route::getCurrentRoute()->getActionName();

You can find everything about laravel Routes in the Laravel API: http://laravel.com/api/5.0/Illuminate/Routing/Router.html http://laravel.com/api/5.0/Illuminate/Routing.html

您可以在laravel API中找到有关laravel路由的所有内容:http://laravel.com/api/5.0/Illuminate/Routing/Router.html http://laravel.com/api/5.0/Illuminate/Routing.html

Retrieving The Request URI

检索请求URI

The path method returns the request's URI. So, if the incoming request is targeted at http://example.com/foo/bar, the path method will return foo/bar:

路径方法返回请求的URI。因此，如果传入请求的目标是http://example.com/foo/bar，那么path方法将返回foo/bar:

$uri = $request->path();

The is method allows you to verify that the incoming request URI matches a given pattern. You may use the * character as a wildcard when utilizing this method:

is方法允许您验证传入的请求URI是否与给定的模式匹配。使用此方法时，您可以使用*字符作为通配符:

if ($request->is('admin/*')) {
    //
}

To get the full URL, not just the path info, you may use the url method on the request instance:

要获得完整的URL，而不仅仅是路径信息，您可以在请求实例上使用URL方法:

$url = $request->url();

#2

Using Laravel 5.1, you can use

使用Laravel 5.1，您可以使用

\Request::route()->getName()

#3

Found a way to find the current route name works for laravel v5 , v5.1.28 and v5.2.10

找到一种方法来找到laravel v5、v5.1.28和v5.2.10的当前路由名称

Namespace

名称空间

use Illuminate\Support\Facades\Route;

and

和

$currentPath= Route::getFacadeRoot()->current()->uri();

For Laravel laravel v5.3 you can just use:

对于Laravel的v5.3，你可以使用:

use Illuminate\Support\Facades\Route;

Route::currentRouteName();

#4

If you need url, not route name, you do not need to use/require any other classes:

如果您需要url，而不是路由名称，则不需要使用/需要任何其他类:

url()->current();

#5

If you want to select menu on multiple routes you may do like this:

如果你想在多条路线上选择菜单，你可以这样做:

<li class="{{ (Request::is('products/*') || Request::is('products') || Request::is('product/*') ? 'active' : '') }}"><a href="{{url('products')}}"><i class="fa fa-code-fork"></i>&nbsp; Products</a></li>

Or if you want to select just single menu you may simply do like this:

或者，如果你只想选择一个菜单，你可以这样做:

<li class="{{ (Request::is('/users') ? 'active' : '') }}"><a href="{{url('/')}}"><i class="fa fa-envelope"></i>&nbsp; Users</a></li>

Also tested in Laravel 5.2

在Laravel 5.2中也进行了测试

Hope this help someone.

希望这帮助别人。

#6

Laravel 5.2 You can use

Laravel 5.2可以使用

$request->route()->getName()

It will give you current route name.

它会给你当前的路径名。

#7

You can use in template:

你可以在模板中使用:

<?php $path = Route::getCurrentRoute()->getPath(); ?>
<?php if (starts_with($path, 'admin/')) echo "active"; ?>

#8

Shortest way is Route facade \Route::current()->getName()

最短路径是路由facade \路由::current()->getName()

This also works in laravel 5.4.*

这也适用于laravel 5.4 *

#9

Now in Laravel 5.3 I am seeing that can be made similarly you tried:

在Laravel 5。3中，我发现可以做类似的东西

$route = Route::current();

$name = Route::currentRouteName();

$action = Route::currentRouteAction();

https://laravel.com/docs/5.3/routing#accessing-the-current-route

https://laravel.com/docs/5.3/routing accessing-the-current-route

#10

In a controller action, you could just do:

在控制器动作中，你可以做:

public function someAction(Request $request)
{
    $routeName = $request->route()->getName();
}

$request here is resolved by Laravel's service container.

这里的$request由Laravel的服务容器解决。

getName() returns the route name for named routes only, null otherwise (but you could still explore the \Illuminate\Routing\Route object for something else of interest).

getName()只返回已命名路由的路由名称，否则为null(但是您仍然可以为其他感兴趣的东西探索\照明的路由对象)。

In other words, you should have your route defined like this to have "nameOfMyRoute" returned:

换句话说，您应该像这样定义您的路由，使“nameOfMyRoute”返回:

Route::get('my/some-action', [
    'as' => 'nameOfMyRoute',
    'uses' => 'MyController@someAction'
]);

#11

Request::path(); is better, and remember to Use Request;

要求::路径();是更好的，记得使用请求;

#12

In 5.2, you can use the request directly with:

在5.2中，您可以直接使用以下请求:

$request->route()->getName();

or via the helper method:

或透过辅助方法:

request()->route()->getName();

Output example:

输出的例子:

"home.index"

#13

I have used for getting route name in larvel 5.3

我在larvel 5.3中使用过获取路线名称

Request::path()

要求::路径()

#14

In a Helper file,

在一个辅助文件中,

Your can use Route::current()->uri() to get current URL.

您可以使用Route::current()->uri()获取当前URL。

Hence, If you compare your route name to set active class on menu then it would be good if you use

因此，如果您将您的路由名称与在菜单上设置活动类进行比较，那么最好使用

Route::currentRouteName() to get the name of route and compare

路由:currentRouteName()获取路由的名称并进行比较

#15

Looking at \Illuminate\Routing\Router.php you can use the method currentRouteNamed() by injecting a Router in your controller method. For example:

看着\照亮\路由\路由器。通过在控制器方法中注入路由器，您可以使用currentRouteNamed()方法。例如:

use Illuminate\Routing\Router;
public function index(Request $request, Router $router) {
   return view($router->currentRouteNamed('foo') ? 'view1' : 'view2');
}

or using the Route facade:

或使用路线外立面:

public function index(Request $request) {
   return view(\Route::currentRouteNamed('foo') ? 'view1' : 'view2');
}

You could also use the method is() to check if the route is named any of the given parameters, but beware this method uses preg_match() and I've experienced it to cause strange behaviour with dotted route names (like 'foo.bar.done'). There is also the matter of performance around preg_match() which is a big subject in the PHP community.

您也可以使用方法is()来检查路由是否被命名为任何给定的参数，但是要注意，这个方法使用preg_match()，我曾经经历过它会导致带有点路径名的奇怪行为(比如'foo.bar.done')。还有关于preg_match()的性能问题，这是PHP社区中的一个重要主题。

public function index(Request $request) {
    return view(\Route::is('foo', 'bar') ? 'view1' : 'view2');
}

#1

252