通过覆盖旧密码来插入新密码 [英] Inserting new password by overriding old password

问题描述

在yii中，我正在创建项目.验证用户输入的电子邮件后，我正在显示password.php文件，该文件具有用于输入新密码的文本字段. Password.php =

In yii i am creating project. After validation of user's entered email, i am displaying password.php file which is having textfield for entering new password. Password.php=

<?php $form=$this->beginWidget('CActiveForm', array(
    'id'=>'email-form',
    'enableClientValidation'=>true,
   ));
        echo CHtml::textField('Enter new password');
        echo CHtml::textField('Repeat password');
        echo CHtml::submitButton('Submit');
        $this->endWidget();

当用户输入新密码并单击提交"按钮时，我想将此新密码插入用户"表的密码字段中，以使其覆盖旧密码.

When user will enter new password and click on submit button i want to insert this new password into User table's password field, in such a way that it overright old password.

在控制器中，我创建的方法为-

In controller i had created method as-

public function actionCreate(){

if(isset($_POST['email']))
    {

 $record=User2::model()->find(array(
'select'=>'userId, securityQuestionId, primaryEmail',
'condition'=>'primaryEmail=:email',
'params'=>array(':email'=>$_POST['email']))
);

if($record===null) {
                  echo "Email invalid";
                   }

else {
    echo "email exists";


      $this->render('Password');
      if(isset($_POST['Password']))
        {

        $command = Yii::app()->db->createCommand();
        $command->insert('User', array(
                    'password'=>$_POST['password'] ,
                    //'params'=>array(':password'=>$_POST['password'])

        }          



 }

    }
    else{
        $this->render('emailForm'); //show the view with the password field
    }

但未插入新密码.那么我该如何实施...请帮助我

But its not inserting new password. So How can i implement this...Please help me

推荐答案

您无法获得像$_POST['Password']这样的密码，因为您尚未设置此post变量.

You can't get password like this $_POST['Password'], because you haven't set this post variable.

您必须使用:

echo CHtml::textField('password');
echo CHtml::textField('repeatPassword');
echo CHtml::hiddenField('email', $email);

'password'和'repeatPassword'是POST变量的名称

'password' and 'repeatPassword' are names of POST vars

在您的控制器中，您有太多错误，请尝试以下操作(检查输入错误):

And in your controller you have too many mistakes, try this (check for typos):

if(isset($_POST['email']))
{
 $record=User2::model()->find(array(
'select'=>'userId, securityQuestionId, primaryEmail',
'condition'=>'primaryEmail=:email',
'params'=>array(':email'=>$_POST['email']))
);

if($record===null) {
                  echo "Email invalid";
                   }

else {
    if(isset($_POST['password']) && isset($_POST['repeatPassword']) && ($_POST['password'] === $_POST['repeatPassword']))
    {
        $record->password = $_POST['password'];
        if ($record->save()) {
             $this->render('saved');
        }
    }          

    $this->render('Password' array('email'=>$_POST['email']));
}

 }

    }
    else{
        $this->render('emailForm'); //show the view with the password field
    }

在您的代码中if(isset($_POST['Password']))将永远不会执行，因为发送密码后，您尚未设置email变量.所以你只是$this->render('emailForm');.因此，我们通过CHtml::hiddenField('email', $email);

In your code if(isset($_POST['Password'])) won't ever execute, because after sending password you haven't set email variable. So you just $this->render('emailForm');. Thus we set it by CHtml::hiddenField('email', $email);

更新.我强烈建议您阅读本指南 .这样可以为您节省很多时间.

Upd. I strongly recommend you to read this guide. It will save a lot of time for you.

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