使用参数修改 symfony 表单的 url

Question

我想使用参数修改提交的 symfony 表单的 url。尝试了该平台的许多解决方案，none 的解决方案有所帮助！

当前url如下：http://localhost:8000/search?app_bundle_search_form%5Bsearch%5D=qui&app_bundle_search_form%5Bbrand%5D=&app_bundle_search_form%5Bprice%5D=500%2C100000&app_bundle_search_form%5B_token%5D=BtA5bZb9HErUXzXFzGFbpEhlD6nD33zr7tKiPLxjpy4

我想要它像`http://localhost:8000/search?search=qui?brand=?minprice=500?maxprice=100000

这是我的控制器：

 public function searchAction($searchTerm=null,Request $request)
{
    if ($request->getMethod() == 'GET') {

        $searchTerm = $request->query->get('app_bundle_search_form')['search'];
        $searchBrand = $request->query->get('app_bundle_search_form')['brand'];
        $price = $request->query->get('app_bundle_search_form')['price'];
        $price = explode(",", $price);

        $minPrice = $price[0];
        $maxPrice = $price[1];

        $em = $this->getDoctrine()->getManager();
        $search = $em->getRepository('AppBundle:Classified')->searchClassifieds($searchTerm, $searchBrand, $minPrice, $maxPrice);


    }


    return $this->render('search-result.html.twig', [
        'searchTerm' => $searchTerm,
        'results' => $search
    ]);


}

表格：

 public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder
        ->setMethod('GET')
        ->add('search', TextType::class,array(
            'required'=> false
        ))
        ->add('brand',EntityType::class,[
            'class'=>Brand::class,
            'placeholder'=>'Choose a brand',
            'required'=>false,
            'query_builder'=>function(BrandRepository $repo){
                return $repo->DistinctBrandValue();
            }
        ])
        ->add('price', TextType::class, array(
            'required' => true,
            'label' => 'Price',
            'attr' => [
                'data-slider-min' => '500',
                'data-slider-max' => '100000',
                'data-slider-step' => '2',
                'data-slider-value' => "[500,100000]",
            ]))
      ;

}

路由

search:
path: /search
defaults:
    _controller: AppBundle:Search:search
requirements:
        methods: GET

树枝：

 {{ form_start(search, { action: path('search')}) }}
    {{ form_widget(search) }}
<button class="btn btn-outline-success my-2 my-sm-0" type="submit">{% trans %}Search{% endtrans %}</button>
<a href="{{ path('search') }}" class="btn btn-primary">{% trans %}Cancel{% endtrans %}</a>
{{ form_end(search) }}

提前致谢！

编辑： 所以我所做的是我必须将表单名称设为空。

public function getBlockPrefix()
{
    return '';
}

并将 csrf 保护设置为 false。

public function configureOptions(OptionsResolver $resolver)
{
    $resolver->setDefaults([
        'csrf_protection' => false,
    ]);

}

并且还在控制器中将表单名称更改为 null。现在 url 看起来好多了！

Answer 1

您可以覆盖 formType 和 return 中的 getName 方法一个空字符串，例如：

表格：

public function getName()
{
    return '';
}

希望对您有所帮助

使用参数修改 symfony 表单的 url

Modify url of a symfony form using parameters

php

symfony-forms

symfony