php - 基于另一个字段的验证

Question

我是这个Symfony框架的新手，在实施过程中遇到了死胡同。仅当输入用户时，我才需要验证new password和confirm password字段current password。

我尽力通过这些链接来理解这个概念，

• http://shout.setfive.com/2013/06/27/symfony2-forms-without-an-entity-and-with-a-conditional-validator/
• 根据 Symfony 2 中的 fieldA 或 fieldB 验证表单 fieldA
• http://tomislavsantek.iz.hr/2011/03/using-symfony-postvalidator/

但事实证明，所使用的类要么已弃用，要么需要一个实体。

两个字段的实现如下，

//if this field is filled
$builder->add('currentPassword', 'password', array('label'=>'Current Password',                                            
                                      'required'=>false,                                          
                                      'attr'=>array('class'=>'form-control'),                                           
                                      'error_bubbling' => true,
                                      'trim' => true,
                                      'mapped' => false,
                                      'label_attr'=>array('class'=>'col-sm-4 control-label')));

//These repeated fields must be filled or must be set as required
$builder->add( 'password', 'repeated', array( 'type' => 'password',                                          
                                      'required' => false,        
                                      'invalid_message' => ErrorMessages::PASSWORDS_DO_NOT_MATCH,
                                      'options' => array('attr' => array('class' => 'password-field form-control')),                                                                                   
                                      'first_options'  => array('label' => false,  
                                                                'error_bubbling' => true,
                                                                'label_attr'=>array('class'=>'col-sm-4 control-label')),
                                      'second_options' => array('label' => false,                                                                    
                                                                'label_attr'=>array('class'=>'col-sm-4 control-label')))); 

我在控制器中使用一堆 if条件实现了验证，但如果能学习为这样的场景执行验证的正确方法，那就太好了。:)

谢谢

编辑

用户entity

    <?php
    namespace Proj\Bundle\AccountsBundle\Entity;

    use Symfony\Component\Validator\Constraints as Assert;
    use Symfony\Component\Security\Core\User\UserInterface;
    use Proj\Bundle\AccountsBundle\Custom\ErrorMessages;



    class User implements UserInterface, \Serializable {    

      /**
       * @Assert\Email(message=ErrorMessages::EMAIL_ADDRESS_INVALID)
       * @Assert\NotBlank(message=ErrorMessages::EMAIL_ADDRESS_EMPTY)
       */
      private $email;

      /**     
       * @Assert\NotBlank(message=ErrorMessages::PASSWORD_EMPTY, groups={"full"})
       */
      private $password;

      private $oldPassword;

      private $id;
      private $userId;
      private $name;
      private $username;



      public function __construct() {

      } 

      function setEmail ($email) {
        $this->email = $email;
        $this->username = $email;
      }

      function getEmail () {
        return $this->email;
      }

      function setPassword ($password) {
        $this->password = $password;
      }

      function getPassword () {
        return $this->password;
      }

      function setOldPassword ($oldPassword) {
        $this->oldPassword = $oldPassword;
      }

      function getOldPassword () {
        return $this->oldPassword;
      }

      function setId ($id) {
        $this->id = $id;
      }

      function getId () {
        return $this->id;
      }

      function setUserId ($userId) {
        $this->userId = $userId;
      }

      function getUserId () {
        return $this->userId;
      }

      function setName (PersonName $name) {
        $this->name = $name;
      }

      function getName () {
        return $this->name;
      }

      public function eraseCredentials() {

      }

      public function getRoles() {
        return array('ROLE_USER');
      }

      public function getSalt() {

      }

      public function getUsername() {
        return $this->username;
      }

    }

Answer 1

修改你的类如下

use Symfony\Component\Validator\Constraints as Assert;
use Symfony\Component\Validator\ExecutionContext;

/**
 *
 * @Assert\Callback(methods={"passwordVerify"})
 */
class User implements UserInterface, \Serializable {
  //all your old code here
  public function passwordVerify(ExecutionContext $context)
  {
    //your controls about password fields here
    //in case of failure you can add that snippet of code
    $context->addViolationAtPath($propertyPath,'your message here', array(), null);
  }
}

当然，您必须能够将所有信息访问到passwordVerify函数中，最快的方法是在verifyPassword实体中创建字段，这样当您将表单与实体绑定时，所有数据都将在那里。

isValid()当您使用表单的方法时，该代码片段将自动调用回调方法