我正在失去Symfony2中表单字段的错误关联

时间:2011-11-07 19:26:00

标签: php ajax doctrine-orm symfony

我的验证看起来像

Acme\UserBundle\Entity\User:
    constraints:
        - \Symfony\Bridge\Doctrine\Validator\Constraints\UniqueEntity: { fields:username, message: "Username already in use" }
        - \Symfony\Bridge\Doctrine\Validator\Constraints\UniqueEntity: { fields:email, message: "Email address already in use" }
    properties:
        username:
            - NotBlank: ~
            - MinLength: { limit: 2, message: "Your username must have at least {{ limit }} characters." }
        email:
            - Email:
                message: The email "{{ value }}" is not a valid email.
                checkMX: true

我的控制器如:

$form = $this->createForm(new RegistrationType());
$form->bindRequest($request);
if ($form->isValid()) {
    //... save to db
}else{
    $errors = $form->getErrors();
    //... pass the errors back as json
}

我正在尝试构建通过ajax请求提交的用户注册控制器。但是,当触发验证错误时,$error变量如下所示:

[2011-11-07 19:19:44] app.INFO: array (
  0 => 
  Symfony\Component\Form\FormError::__set_state(array(
     'messageTemplate' => 'Email address already in use',
     'messageParameters' => 
    array (
    ),
  )),
  1 => 
  Symfony\Component\Form\FormError::__set_state(array(
     'messageTemplate' => 'Your username must have at least {{ limit }} characters.',
     'messageParameters' => 
    array (
      '{{ value }}' => '1',
      '{{ limit }}' => 2,
    ),
  )),
) [] []

问题是我不知道错误对应哪个字段。有没有办法找到这些数据,这样当我发送json响应时,我可以将错误消息与相关字段相关联。

2 个答案:

答案 0 :(得分:6)

我认为您可以单独查询每个字段,例如:

$form->get('username')->getErrors()

所以你可以这样建立一个数组:

$errors['username'] = $form->get('username')->getErrors();
$errors['email'] = $form->get('email')->getErrors();

您可以通过以下方式实现自动化:

$fields = $form->getChildren();
foreach ( $fields as $field ) {
    $errors[$field->getName()] = $field->getErrors();
}

我认为getName函数应该在自己的表单的子项上调用时返回字段名称。其他人可能会有更有效的方式......

答案 1 :(得分:3)

我认为已接受的答案已经过时,不再有SELECT tablesite.name, tablesite.family, tablesite.phone_number, COUNT(action.service_provider_id) FROM tablesite LEFT JOIN action ON tablesite.id_user=action.service_provider_id AND action.vote !='' AND action.customer_comment ='' GROUP BY name, family, phone_number 方法了。

getChildren()

最后,如果您想要从生成的$errors = []; foreach ($form->all() as $field) { if ($field->getErrors()->count() > 0) { $fieldName = $field->getName(); $errors[$fieldName] = []; foreach ($field->getErrors() as $error) { $errors[$fieldName][] = $error->getMessage(); } } } 访问表单字段,它会为您提供完整的输入字段名称:

FormView