我有一個日誌的形式,檢查通過數據庫的用戶電子郵件地址和密碼,如果匹配,它允許用戶登錄。問題是,它會檢查數據庫匹配任何密碼的電子郵件或與數據庫中任何電子郵件相匹配的密碼。我希望它檢查這個特定的用戶電子郵件以匹配他的密碼,不匹配數據庫中存在的任何密碼。Zend框架的登錄表單
這裏是我的控制器,我認爲我沒有錯:
$loginForm = new Application_Form_UserLogin();
if ($this->getRequest()->isPost('loginForm'))
{
$email_adrress = $this->getRequest()->getParam('email_address');
$password = $this->getRequest()->getParam('password');
/************ Login Form ************/
if ($loginForm->isValid($this->getRequest()->getParams()))
{
$user = $this->_helper->model('Users')->createRow($loginForm->getValues());
$user = $this->_helper->model('Users')->fetchRowByFields(array('email' => $email_adrress, 'hash' => $password));
if($user)
{
Zend_Session::rememberMe(86400 * 14);
Zend_Auth::getInstance()->getStorage()->write($user);
$this->getHelper('redirector')->gotoRoute(array(), 'invite');
return;
}
else {
}
}
}$this->view->loginForm = $loginForm;
我的形式:
class Application_Form_UserLogin extends Zend_Form
{
public $email, $password, $submit;
public function init()
{
$this->setName('loginForm');
$EmailExists = new Zend_Validate_Db_RecordExists(
array(
'table' => 'users',
'field' => 'email'
)
);
//$EmailExists->setMessage('Invalid email address, please try again. *');
$PasswordExists = new Zend_Validate_Db_RecordExists(
array(
'table' => 'users',
'field' => 'hash'
)
);
$PasswordExists->setMessage('Invalid password, please try again. *');
$this->email = $this->createElement('text', 'email_address')
->setLabel('Email')
->addValidator($EmailExists)
->addValidator('EmailAddress')
->setRequired(true);
$this->password = $this->createElement('text', 'password')
->setLabel('Password')
->addValidator($PasswordExists)
->setRequired(true);
$this->submitButton = $this->createElement('button', 'btn_login')
->setLabel('Login')
->setAttrib('type', 'submit');
$this->addElements(array($this->email, $this->password, $this->submit));
$elementDecorators = array(
'ViewHelper'
);
$this->setElementDecorators($elementDecorators);
}
}
哦,謝謝我會試試看。我正在使用Zend_Validate_Db_RecordExists來驗證登錄... –