我需要在Yii中以登录形式散列并存储来自用户输入的密码。 如果我通过这样的POST参数得到它们:
$model->username=$_POST['User']['username'];
$model->password=crypt($_POST['User']['username']);// salt might be added
if($model->save())
$this->redirect(array('view','id'=>$model->id));
这样我在POST请求中公开了未加密的密码。 其他方式是直接从登录表单这样:
public function actionCreate2()
{
$model=new User;
$model->username = $form->username;
$model->password = crypt($form->password);
if($model->save())
$this->redirect(array('view','id'=>$model->id));
$this->render('create',array(
'model'=>$model,
));
}
但在我的情况下,无效验证已保存的用户。 auth功能:
public function authenticate()
{
$users = User::model()->findByAttributes(array('username'=>$this->username));
if($users == null)
$this->errorCode=self::ERROR_USERNAME_INVALID;
elseif ($users->password !== crypt($this->password, $users->password))
//elseif($users->password !== $this->password)
$this->errorCode=self::ERROR_PASSWORD_INVALID;
else
$this->errorCode=self::ERROR_NONE;
return !$this->errorCode;
}
如何以正确的方式做到这一点?
随着我跟随Samuel的出现,出现了更多麻烦 - 甚至在我输入任何内容之前,验证警报消息,以及输入字段中的散列密码。(见图):
当我仍然输入我的用户名和密码而不是“建议”并按“创建”时,表单将被发送时没有加密值(来自POST请求嗅探):
Form Data view source view URL encoded
YII_CSRF_TOKEN:9758c50299b9d4b96b6ac6a2e5f0c939eae46abe
User[username]:igor23
User[password]:igor23
yt0:Create
但实际上没有任何内容存储在db中,也没有加密未加密...
答案 0 :(得分:1)
将您的创建方法更改为:
/**
* Creates a new model.
* If creation is successful, the browser will be redirected to the 'view' page.
*/
public function actionCreate() {
$model = new User;
if (isset($_POST['User'])) {
$model->attributes = $_POST['User'];
$model->password = crypt($model->password, 'mysalt123');
if ($model->save())
$this->redirect(array('view', 'id' => $model->primaryKey));
}
// Reset password field
$model->password = "";
$this->render('create', array(
'model' => $model,
));
}
从那里改变那个:
elseif ($users->password !== crypt($this->password, $users->password))
对此:
elseif (strcmp(crypt($this->password, 'mysalt123'), $users->password))