登录我的应用程序时遇到问题。每当我点击提交按钮时,我都会收到此消息:
登录无效。
我认为问题出在我的模型上,但似乎无法找到问题。我想要的是验证名为employee_login的表中的用户名和密码,我的数据库是Employee
控制器:
public function index(){
echo "<script> document.location.href='".$this->config->base_url()."index.php/admin/login' </script>";
}
public function login()
{
$this->load->helper('url');
$this->load->model('admin_model');
$this->load->database('default', TRUE);
if(isset($_POST['submit'])){
$this->admin_model->validate();
}
$_SESSION['search_text']="";
$this->load->view('login');
型号:
function get_data($table, $fields, $condition,$orderby=NULL) {
$this->load->database();
$condition = (isset($condition))? ' WHERE '.$condition : '';
$orderby = (isset($orderby))? ' ORDER BY '.$orderby : '';
//if(!$condition) $condition = 'active = A';
//if(!$orderby) $orderby = 'id DESC';
$result = $this->db->query('SELECT '.$fields.' FROM '.$table.$condition.$orderby );
return $result;
}
function validate() {
$username=$this->input->post('Username');
$pass=$this->input->post('Password');
$this->load->database();
$res=$this->db->query("SELECT count(*) AS cnt,id,emp_type,employee_name, report_to FROM employee_login WHERE username='$username' AND password='".md5($pass)."'");
foreach ($res->result() as $rows){
$cnt=$rows->cnt;
$uid=$rows->id;
$emp_type=$rows->emp_type;
$report_to=$rows->report_to;
$emp_name=$rows->employee_name;
}
if($cnt>0){
$_SESSION['DATABASE_uid']=$uid;
$_SESSION['emp_type']=$emp_type;
$_SESSION['report_to']=$report_to;
$_SESSION['emp_name']=$emp_name;
if($_SESSION['DATABASE_uid']==1 || $_SESSION['DATABASE_uid']==5)
echo "<script> document.location.href='../admin/leave_listing' </script>";
else
echo "<script> document.location.href='../admin/apply_leave' </script>";
}
else
echo "<script> alert('Invalid Login') </script>";
}
数据库:
CREATE TABLE IF NOT EXISTS `employee_login` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`employee_name` varchar(500) NOT NULL,
`username` varchar(500) NOT NULL,
`password` varchar(500) NOT NULL,
`emp_type` varchar(50) NOT NULL,
`emp_email` varchar(100) NOT NULL,
`report_to` varchar(100) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=4 ;