在 Twig 中显示 flash 消息的正确方法

Question

我有一个注册表，提交后，显示快速消息并重定向到特定 pages.My 如果没有错误，快速成功消息工作正常 found.However，错误消息不会显示，当有错误示例空白字段，而是显示默认的 Symfony2 异常。

(DBALexception, PDOexception,SQLexceoption,etc) [2/2] DBALException: An exception occurred while executing 'INSERT INTO voters ( blah and blah......)

这是在开发阶段，我想测试错误信息，我想再次显示表单，错误闪烁而不是 Symfony2 500 错误页面；

如何在特定页面中暂时禁用 Symfony2 异常，而不是在开发过程中显示错误提示信息？

我在控制器里有这个

if ($form->isValid()) {
        $em = $this->getDoctrine()->getManager();
        $em->persist($entity);
        $em->flush();


        $this->addFlash('notice', 'Welcome to the growing lists of Supporters, dont forget to share and invite this to your friends and relatives, have a magical day!');

        //return $this->redirect($this->generateUrl('vo_show', array('id' => $entity->getId())));
        return $this->redirect($this->generateUrl('vo'));
    }
    else {

        $this->addFlash('error', 'Welcome to the Death Star, have a magical day!');

        return $this->render('Bundle:Vo:new.html.twig', array(
        'entity' => $entity,
        'form'   => $form->createView(),
    ));

在树枝上

{% if app.session.flashBag.has('notice') %}
        <div class="alert alert-success fade in" role="alert">
            <a href="#" class="close" data-dismiss="alert" aria-label="close">&times;</a>
            {% for msg in app.session.flashBag.get('notice') %}
                {{ msg }}
            {% endfor %}
        </div>
    {% endif %}
    {% if app.session.flashBag.has('error') %}
        <div class="alert alert-error fade in" role="alert">
            <a href="#" class="close" data-dismiss="alert" aria-label="close">&times;</a>
            {% for msg in app.session.flashBag.get('error') %}
                {{ msg }}
            {% endfor %}
        </div>
    {% endif %}


//registerform.twig.html

{{ form_start(form, {attr: {novalidate: 'novalidate'}} ) }}
        {{ form_errors(form) }}
        {{ form_row(form.comments,{'attr': {'placeholder': 'Why You Want Death'}}) }}
        {{ form_end(form) }}

所以我不想显示 Symfony2 异常，而是想重定向回表单并在表单提交失败时显示个别表单错误

在 Symfony 1.4 中，我可以很容易地在 form.class 中做到这一点，它扩展了基础 class，就像这样

$this->mergePostValidator(new sfValidatorDoctrineUnique(array(
  'model' => 'Voters',
  'column' => array('firstname', 'middlename', 'lastname', 'city_id', 'birthday', 'profession_id')),array('invalid' => '<div class="alert alert-warning">You are already registered.No need to proceed buddy</div>')
));

如果填写表单时出现错误，它会在网页中创建一个 flash 错误消息并重新显示表单而不是重定向到 501 page.Otherwise 它会创建一个成功的 flash message.I想以相同的方式进行 html5 验证，其中每个错误都以表单

呈现

Answer 1

你不能那样做。异常'break'程序的流程，遇到异常不能简单的继续执行代码。
但是，您可以手动捕获代码中的异常，然后像这样显示出错的地方：

if ($form->isValid()) {
    try{
    $em = $this->getDoctrine()->getManager();
    $em->persist($entity);
    $em->flush();
    } catch(\Exception $e){
        $this->addFlash('notice', sprintf('A %s was thrown when trying to persist 
                      the entities with message = %s', get_class($e), $e->getMessage());
   }
// Rest of your code ...
// Take note that when you encounter exception the flow of the program
// is 'broken' and you must carefully examine what happened to avoid 
// unintended consequences and stuff like null valued variables

Answer 2

创建 YML 格式的约束条件

例如

Project\Bundle\YourBundle\Entity\Vo:
properties:
    firstname:
        - NotBlank: ~

在 Twig 中显示 flash 消息的正确方法

Correct way of displaying flash messages in Twig

symfony

twig