【发布时间】:2014-04-04 02:54:42
【问题描述】:
有没有办法在 Symfony2 控制器中获取动作的名称?
public function createAction(Request $request, $title) {
// Expected result: create
$name = $this->getActionName();
}
【问题讨论】:
标签: php symfony controller action
有没有办法在 Symfony2 控制器中获取动作的名称?
public function createAction(Request $request, $title) {
// Expected result: create
$name = $this->getActionName();
}
【问题讨论】:
标签: php symfony controller action
使用:
$request->attributes->get('_controller');
// will get yourBundle\Controller\yourController::CreateAction
$params = explode('::',$request->attributes->get('_controller'));
// $params[1] = 'createAction';
$actionName = substr($params[1],0,-6);
// $actionName = 'create';
【讨论】:
我找到了这个 sn-p (here):
$matches = array();
$controller = $this->getRequest()->attributes->get('_controller');
preg_match('/(.*)\\\(.*)Bundle\\\Controller\\\(.*)Controller::(.*)Action/', $controller, $matches);
这似乎是一个很有前途的方法。 这个正则表达式实际上不起作用。但是使用 Works!strstr()获取动作名称并不难。
然后返回 (see example)
Array
(
[0] => Acme\MyBundle\Controller\MyController::myAction
[1] => Acme
[2] => My
[3] => My
[4] => my
)
如果输入是Acme\MyBundle\Controller\MyController::myAction。
【讨论】:
现在,我将它与 Symfony 2.8(和 Symfony3)一起使用:
<?php
namespace Company\Bundle\AppBundle\Component\HttpFoundation;
use Symfony\Component\HttpFoundation\Request as BaseRequest;
/**
* Request shortcuts.
*/
class Request extends BaseRequest
{
/**
* Extract the action name.
*
* @return string
*/
public function getActionName()
{
$action = $this->get('_controller');
$action = explode('::', $action);
// use this line if you want to remove the trailing "Action" string
//return isset($action[1]) ? preg_replace('/Action$/', '', $action[1]) : false;
return $action[1];
}
/**
* Extract the controller name (only for the master request).
*
* @return string
*/
public function getControllerName()
{
$controller = $this->get('_controller');
$controller = explode('::', $controller);
$controller = explode('\\', $controller[0]);
// use this line if you want to remove the trailing "Controller" string
//return isset($controller[4]) ? preg_replace('/Controller$/', '', $controller[4]) : false;
return isset($controller[4]) ? $controller[4] : false;
}
}
要使用此自定义请求类,您必须在您的 web/app*.php 控制器中“使用”它:
use Company\Bundle\AppBundle\Component\HttpFoundation\Request;
// ...
$request = Request::createFromGlobals();
// ...
然后在你的控制器中:
class AppController extends Controller
{
/**
* @Route("/", name="home_page")
* @Template("")
*
* @return array
*/
public function homePageAction(Request $request)
{
$controllerName = $request->getControllerName();
$actionName = $request->getActionName();
dump($controllerName, $actionName); die();
// ...
}
将输出:
AppController.php on line 27:
"AppController"
AppController.php on line 27:
"homePageAction"
您也可以通过RequestStack服务访问这些功能:
class MyService
{
/**
* @param RequestStack $requestStack
*/
public function __construct(RequestStack $requestStack)
{
$this->requestStack = $requestStack;
}
public function myService()
{
$this->controllerName = $this->requestStack->getMasterRequest()->getControllerName();
$this->actionName = $this->requestStack->getMasterRequest()->getActionName();
// ...
}
【讨论】:
如果您使用 Controller as a Service,那么架构会有所不同:
$request->attributes->get('_controller'); 将返回“service_id:createAction”
两种模式的可能解决方案:
$controller = $request->attributes->get('_controller');
$controller = str_replace('::', ':', $controller);
list($controller, $action) = explode(':', $controller);
【讨论】:
在所有版本的 symfony 中,没有 $request 或容器、服务或其他任何东西......,直接在你的方法中
public function myMethod(){
$methodName = __METHOD__;
return $methodName;
}
// return App\Controller\DefaultController::myMethod
public function mySecondMethod(){
$methodName = explode('::', __METHOD__);
return $methodName[1];
}
// return mySecondMethod
【讨论】: