如何在 Laravel 4 中的 @if 语句（刀片）中获取当前 URL？
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How to Get the Current URL Inside @if Statement (Blade) in Laravel 4?
提问by user2443854
I am using Laravel 4. I would like to access the current URL inside an @ifcondition in a view using the Laravel's Blade templating engine but I don't know how to do it.
我正在使用 Laravel 4。我想@if使用 Laravel 的 Blade 模板引擎访问视图中条件内的当前 URL，但我不知道该怎么做。
I know that it can be done using something like <?php echo URL::current(); ?>but It's not possible inside an @ifblade statement.
我知道它可以使用类似的东西来完成，<?php echo URL::current(); ?>但在@if刀片语句中是不可能的。
Any suggestions?
有什么建议？
回答by Andreyco
You can use: Request::url()to obtain the current URL, here is an example:
您可以使用：Request::url()来获取当前的 URL，这里是一个例子：
@if(Request::url() === 'your url here')
// code
@endif
Laravel offers a method to find out, whether the URL matches a pattern or not
Laravel 提供了一种方法来确定 URL 是否与模式匹配
if (Request::is('admin/*'))
{
// code
}
Check the related documentation to obtain different request information: http://laravel.com/docs/requests#request-information
查看相关文档获取不同的请求信息：http: //laravel.com/docs/requests#request-information
回答by Naing Win
You can also use Route::current()->getName()to check your route name.
您还可以使用Route::current()->getName()来检查您的路线名称。
Example: routes.php
示例：routes.php
Route::get('test', ['as'=>'testing', function() {
return View::make('test');
}]);
View:
看法：
@if(Route::current()->getName() == 'testing')
Hello This is testing
@endif
回答by Dean Gite
Maybe you should try this:
也许你应该试试这个：
<li class="{{ Request::is('admin/dashboard') ? 'active' : '' }}">Dashboard</li>
回答by Jucaman
I'd do it this way:
我会这样做：
@if (Request::path() == '/view')
// code
@endif
where '/view' is view name in routes.php.
其中 '/view' 是 routes.php 中的视图名称。
回答by Sagar Naliyapara
To get current urlin bladeview you can use following,
为了获得current url在blade视图中，可以使用以下，
<a href="{{url()->current()}}">Current Url</a>
So as you can compare using following code,
因此，您可以使用以下代码进行比较，
@if (url()->current() == 'you url')
//stuff you want to perform
@endif
回答by Abduhafiz
This is helped to me for bootstrap active nav class in Laravel 5.2:
这对我在Laravel 5.2 中引导活动导航类有帮助：
<li class="{{ Request::path() == '/' ? 'active' : '' }}"><a href="/">Home</a></li>
<li class="{{ Request::path() == 'about' ? 'active' : '' }}"><a href="/about">About</a></li>
回答by VineFreeman
A little old but this works in L5:
有点旧，但这适用于 L5：
<li class="{{ Request::is('mycategory/', '*') ? 'active' : ''}}">
This captures both /mycategory and /mycategory/slug
这会同时捕获 /mycategory 和 /mycategory/slug
回答by Rob
Laravel 5.4
Laravel 5.4
Global functions
全局函数
@if (request()->is('/'))
<p>Is homepage</p>
@endif
回答by Alias
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 $url = Request::url();.
我个人不会尝试在视图中抓取它。我对 Laravel 并不感兴趣，但我想你需要将路由发送到控制器，然后在控制器中，使用类似$url = Request::url();.
One way of doing it anyway.
无论如何，一种方法。
EDIT: Actually look at the method above, probably a better way.
编辑：实际上看看上面的方法，可能是更好的方法。
回答by Alejandro Silva
A simple navbar with bootstrap can be done as:
一个带有引导程序的简单导航栏可以这样完成：
<li class="{{ Request::is('user/profile')? 'active': '' }}">
<a href="{{ url('user/profile') }}">Profile </a>
</li>
