Php yii用户标识错误代码

但是，在我看来，它返回的是错误的用户名或密码，而不是错误的非活动消息。我应该如何纠正此错误？

在AUAuthentication函数中的

LoginForm

模型中，您需要根据错误代码添加错误

class UserIdentity extends CUserIdentity
{
    const ERROR_USERNAME_INACTIVE=67;
    private $_id;

    public function authenticate()
    {
        $username=strtolower($this->username);
        $user=User::model()->find('LOWER(username)=?',array($username));
        if($user===null)
            $this->errorCode=self::ERROR_USERNAME_INVALID;
        else if(!$user->validatePassword($this->password))
            $this->errorCode=self::ERROR_PASSWORD_INVALID;
        else if($user->active == 0)
            $this->errorCode=self::ERROR_USERNAME_INACTIVE; 
        else
        {
            $this->_id=$user->id;
            $this->username=$user->username;
            $this->errorCode=self::ERROR_NONE;
        }
        return $this->errorCode==self::ERROR_NONE;
    }

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

---

在AUAuthenticate函数中的

LoginForm

模型中，需要根据错误代码添加错误

class UserIdentity extends CUserIdentity
{
    const ERROR_USERNAME_INACTIVE=67;
    private $_id;

    public function authenticate()
    {
        $username=strtolower($this->username);
        $user=User::model()->find('LOWER(username)=?',array($username));
        if($user===null)
            $this->errorCode=self::ERROR_USERNAME_INVALID;
        else if(!$user->validatePassword($this->password))
            $this->errorCode=self::ERROR_PASSWORD_INVALID;
        else if($user->active == 0)
            $this->errorCode=self::ERROR_USERNAME_INACTIVE; 
        else
        {
            $this->_id=$user->id;
            $this->username=$user->username;
            $this->errorCode=self::ERROR_NONE;
        }
        return $this->errorCode==self::ERROR_NONE;
    }

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