基於另一個字段的驗證

Question

我對這個Symfony框架不熟悉，並且在實施過程中遇到了死衚衕。僅當輸入用戶的current password時，我才需要驗證new password和confirm password字段。

我盡力通過去，雖然這些鏈接理解的概念，

• http://shout.setfive.com/2013/06/27/symfony2-forms-without-an-entity-and-with-a-conditional-validator/
• Validate form fieldA based on either fieldA or fieldB in Symfony 2
• http://tomislavsantek.iz.hr/2011/03/using-symfony-postvalidator/

但事實證明無論是所使用的類已棄用或要求一個實體。

兩個領域的實施情況如下，

//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')))); 

我使用控制器內的一堆if條件實施了驗證，但它會是巨大的，學習的情況進行驗證的正確方法如此。 :)

謝謝

編輯

用戶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

修改類如下

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); 
    } 
} 

當然，你必須能夠訪問所有信息進入passwordVerify函數，最快的方法是創建字段verifyPassword到你的實體中，所以當你將form與實體綁定時，所有的數據都會在那裏。

這代碼片段會自動調用當您使用isValid()窗體的方法