php - symfony2 : error when rendering template - Object could not be converted to string

Question

我有这个简单的 Controller :

class SearchController extends Controller{
public function indexAction(Request $request)
{
    $search = new Search();
    $form = $this->createForm(new SearchType(), $search);
    $form->handleRequest($request);

    if ($form->isValid()) {
        return $this->redirect($this->generateUrl('search'));
    }

    return $this->render('MyApplicationBundle:Search:index.html.twig', array(
        'form' => $form->createView(),
    ));
}
}

这是搜索实体:

class Search {
protected $query;

public function setQuery($query)
{
    $this->query = $query;
}

public function getQuery()
{
    return $this->query;
}
}

这是我的表格:

class SearchType extends AbstractType{
public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder
        ->add('query', 'text')
        ->add('search', 'submit');
}

public function getName()
{
    return 'search';
}
}

不幸的是在尝试渲染表单时

{% extends '::base.html.twig' %}
{% block body %}
{{ form(form) }}
{% endblock %}

我遇到了这个错误:

在呈现模板期间抛出异常(“可捕获的 fatal error :类 My\ApplicationBundle\Entity\Search 的对象无法转换为字符串

顺便说一句，只渲染 HTML 效果很好。

有什么想法吗？谢谢你的帮助

已解决:我找到了一种解决方法来更改呈现表单的方式:

    {{ form_start(form, {'attr': {'class': 'form-search'}}) }}
{{ form_widget(form.query, {'attr': {'class': 'form-control search-query'}}) }}
{{ form_end(form) }}

Answer 1

我想我已经找到了您问题的答案 here .

the method inherited from AbstractType will create the name according to the class name which will lead to search. But this will cause issues when rendering as there is a block to render the type search but for the core type. You should set the name explicitly by using a name not used already (a good way is to prefix it by the alias of the bundle)

所以问题可能出在名称 search 本身。然后尝试指定不同的名称。