在Symfony2中,路由参数可以自动映射到控制器参数,例如:http://a.com/test/foo将返回"foo“
/**
* @Route("/test/{name}")
*/
public function action(Request $request, $name) {
return new Response(print_r($name, true));
}请参阅http://symfony.com/doc/current/book/routing.html#route-parameters-and-controller-arguments
但是我想使用查询字符串代替例如:
怎么做?对我来说,只有3种解决方案:
还有别的解决办法吗?
发布于 2022-01-26 13:09:14
与#2一样,要解决私有方法(getIdentifier),首先设置属性并正常执行(父方法::apply)。在Symfony 4.4上测试
<?php
namespace App\FrameworkExtra\Converters;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\ParamConverter;
use Sensio\Bundle\FrameworkExtraBundle\Request\ParamConverter\DoctrineParamConverter;
use Symfony\Component\HttpFoundation\Request;
class QueryStringEntityConverter extends DoctrineParamConverter
{
public function supports(ParamConverter $configuration)
{
return 'querystringentity' == $configuration->getConverter();
}
public function apply(Request $request, ParamConverter $configuration)
{
$param = $configuration->getName();
if (!$request->query->has($param)) {
return false;
}
$value = $request->query->get($param);
$request->attributes->set($param, $value);
return parent::apply($request, $configuration);
}
}https://stackoverflow.com/questions/33982299
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