从 url、laravel 获取控制器动作
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Get controller action from url, laravel
提问by aayush shrestha
Can I get the controller action from given URL?
我可以从给定的 URL 获取控制器操作吗？
In my project, I will have different layout used for admin and normal users. i.e.
在我的项目中，我将为管理员和普通用户使用不同的布局。IE
something.com/content/list - will show layout 1.
something.com/admin/content/list - will show layout 2.
(But these need to be generated by the same controller)
something.com/content/list - 将显示布局 1。
something.com/admin/content/list - 将显示布局 2。
（但这些需要由同一个控制器生成）
I have added filter to detect the pattern 'admin/*'for this purpose. Now I need to call the action required by the rest of the URL ('content/list' or anything that will appear there). Meaning, there could be anything after admin/it could be foo/1/edit(in which case foo controller should be called) or it could be bar/1/edit(in which case bar controller should be called). That is why the controller name should be generated dynamically from the url that the filter captures,
'admin/*'为此，我添加了过滤器来检测模式。现在我需要调用 URL 的其余部分（“ content/list”或将出现在那里的任何内容）所需的操作。意思是，在admin/它之后可能有任何东西foo/1/edit（在这种情况下应该调用 foo 控制器）或者它可能是bar/1/edit（在这种情况下应该调用 bar 控制器）。这就是为什么应该从过滤器捕获的 url 动态生成控制器名称的原因，
So, I want to get the controller action from the URL (content/list) and then call that controller action from inside the filter.
所以，我想从 URL（内容/列表）中获取控制器操作，然后从过滤器内部调用该控制器操作。
Can this be done?
这能做到吗？
回答by aayush shrestha
Thanks to everyone who participated.
感谢所有参加的人。
I just found the solution to my problem in another thread. HERE
我刚刚在另一个线程中找到了解决我的问题的方法。这里
This is what I did.
这就是我所做的。
if(Request::is('admin/*')) {
$my_route = str_replace(URL::to('admin'),"",Request::url());
$request = Request::create($my_route);
return Route::dispatch($request)->getContent();
}
I could not find these methods in the documentation. So I hope, this will help others too.
我在文档中找不到这些方法。所以我希望，这也能帮助其他人。
回答by egig
You can use RESTful Controller
您可以使用RESTful 控制器
Route:controller('/', 'Namespace\yourController');
But the method have to be prefixed by HTTP verb and I am not sure whether it can contain more url segment, in your case, I suggest just use:
但是该方法必须以 HTTP 动词为前缀，我不确定它是否可以包含更多 url 段，在您的情况下，我建议只使用：
Route::group(array('prefix' => 'admin'), function()
{
//map certain path to certain controller, and just throw 404 if no matching route
//it's good practice
Route::('content/list', 'yourController@yourMethod');
});
回答by Anil Saini
Use this in your controller function -
在您的控制器功能中使用它 -
if (Request::is('admin/*'))
{
//layout for admin (layout 2)
}else{
//normal layout (layout 1)
}
回答by The Alpha
You can use Request::segment(index)to get part/segment of the url
您可以使用Request::segment(index)获取部分/段url
// http://www.somedomain.com/somecontroller/someaction/param1/param2
$controller = Request::segment(1); // somecontroller
$action = Request::segment(2); // someaction
$param1 = Request::segment(3); // param1
$param2 = Request::segment(3); // param2
