php 在 Symfony 2 中获取所有请求参数
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Getting all request parameters in Symfony 2
提问by ContextSwitch
In symfony 2 controllers, every time I want to get a value from post I need to run:
在 symfony 2 控制器中，每次我想从 post 获取值时，我都需要运行：
$this->getRequest()->get('value1');
$this->getRequest()->get('value2');
Is there any way to consolidate these into one statement that would return an array? Something like Zend's getParams()?
有没有办法将这些合并为一个返回数组的语句？Zend 的 getParams() 之类的东西？
回答by Guillaume Flandre
You can do $this->getRequest()->query->all();to get all GET params and $this->getRequest()->request->all();to get all POST params.
您可以$this->getRequest()->query->all();获取所有 GET 参数并$this->getRequest()->request->all();获取所有 POST 参数。
So in your case:
所以在你的情况下：
$params = $this->getRequest()->request->all();
$params['value1'];
$params['value2'];
For more info about the Request class, see http://api.symfony.com/2.8/Symfony/Component/HttpFoundation/Request.html
有关 Request 类的更多信息，请参阅http://api.symfony.com/2.8/Symfony/Component/HttpFoundation/Request.html
回答by Aftab Naveed
With Recent Symfony 2.6+ versions as a best practice Request is passed as an argument with action in that case you won't need to explicitly call $this->getRequest(), but rather call $request->request->all()
使用最近的 Symfony 2.6+ 版本作为最佳实践请求作为参数传递在这种情况下，您不需要显式调用 $this->getRequest()，而是调用 $request->request->all()
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Template;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\HttpKernel\Exception\BadRequestHttpException;
use Symfony\Component\HttpKernel\Exception\NotAcceptableHttpException;
use Symfony\Component\HttpFoundation\RedirectResponse;
class SampleController extends Controller
{
public function indexAction(Request $request) {
var_dump($request->request->all());
}
}
