Symfony2：在控制器参数中自动映射查询字符串

Question

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 will return "foo"

    /**
     * @Route("/test/{name}")
     */
    public function action(Request $request, $name) {
        return new Response(print_r($name, true));
    }

见http://symfony.com/doc/current/book/routing.html#route-parameters-and-controller-arguments

但我想改用 查询字符串 例如：http://a.com/test?name=foo

如何做到这一点？ 对我来说只有 3 个解决方案：

• 重新实现ControllerResolverInterface
• 使用自定义 ParamConverter
• $name = $request->query->get('name');

还有其他解决方案吗？

Answer 1

我为那些想要使用转换器的人提供代码：

use Sensio\Bundle\FrameworkExtraBundle\Configuration\ParamConverter;
use Sensio\Bundle\FrameworkExtraBundle\Request\ParamConverter\ParamConverterInterface;
use Symfony\Component\HttpFoundation\Request;

/**
 * Put specific attribute parameter to query parameters
 */
class QueryStringConverter implements ParamConverterInterface{
    public function supports(ParamConverter $configuration) {
        return 'querystring' == $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);
    }
}

services.yml：

services:
  querystring_paramconverter:
    class: AppBundle\Extension\QueryStringConverter
    tags:
      - { name: request.param_converter, converter: querystring }

在您的控制器中：

/**
 * @Route("/test")
 * @ParamConverter("name", converter="querystring")
 */
public function action(Request $request, $name) {
  return new Response(print_r($name, true));
}

Answer 2

基于 Remy 的答案的改进解决方案，它将参数映射到实体：

<?php
namespace AppBundle\Extension;

use Sensio\Bundle\FrameworkExtraBundle\Configuration\ParamConverter;
use Symfony\Component\HttpFoundation\Request;
use Sensio\Bundle\FrameworkExtraBundle\Request\ParamConverter\DoctrineParamConverter;

/**
 * Put specific attribute parameter to query parameters
 */
class QueryStringConverter extends DoctrineParamConverter {

    protected function getIdentifier(Request $request, $options, $name)
    {
        if ($request->query->has($name)) {
            return $request->query->get($name);
        }

        return false;
    }

}

services.yml：

services:
  querystring_paramconverter:
    class: MBS\AppBundle\Extension\QueryStringConverter
    arguments: ['@doctrine']
    tags:
      - { name: request.param_converter, converter: querystring }

在您的控制器中：

/**
 * @Route("/test")
 * @ParamConverter("myobject")
 */
public function action(Request $request, AnyEntity $myobject) {
  return new Response(print_r($myobject->getName(), true));
}

Answer 3

和#2一样，解决私有方法（getIdentifier）首先设置属性并正常执行（parent::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);
    }
}

Answer 4

我没有检查，但似乎 FOSRestBundle 提供了 @QueryParam 注释，它可以做到这一点： http://symfony.com/doc/current/bundles/FOSRestBundle/param_fetcher_listener.html