我儘量去實現簡單的登錄應用程序,我得到這個笨錯誤,但是當我插入的電子郵件和密碼登錄形式,它以相同的形式作出迴應,我無法登錄笨簡單登錄錯誤
這是我的模型功能:
class User_model extends CI_Model {
public function __construct()
{
parent::__construct();
}
function login($email,$password)
{
$this->db->where("email",$email);
$this->db->where("password",$password);
$query=$this->db->get("user");
if($query->num_rows()>0)
{
foreach($query->result() as $rows)
{
//add all data to session
$newdata = array(
'user_id' => $rows->id,
'user_name' => $rows->username,
'user_email' => $rows->email,
'logged_in' => TRUE,
);
}
$this->session->set_userdata($newdata);
return true;
}
return false;
}
控制器:
class User extends CI_Controller{
public function __construct()
{
parent::__construct();
$this->load->model('user_model');
}
public function index()
{
if(($this->session->userdata('user_name')!=""))
{
$this->welcome();
}
else{
$data['title']= 'Home';
$this->load->view('header_view',$data);
$this->load->view("registration_view.php", $data);
$this->load->view('footer_view',$data);
}
}
public function welcome()
{
$data['title']= 'Welcome';
$this->load->view('header_view',$data);
$this->load->view('welcome_view.php', $data);
$this->load->view('footer_view',$data);
}
public function login()
{
$email=$this->input->post('email');
$password=md5($this->input->post('pass'));
$result=$this->user_model->login($email,$password);
if($result) $this->welcome();
else $this->index();
}
查看:
<div id="content">
<div class="signup_wrap">
<div class="signin_form">
<?php echo form_open("user/login"); ?>
<label for="email">Email:</label>
<input type="text" id="email" name="email" value="" />
<label for="password">Password:</label>
<input type="password" id="pass" name="pass" value="" />
<input type="submit" class="" value="Sign in" />
您確定密碼未加密? – sulmanpucit
會發生什麼?你嘗試了什麼,結果如何?當你迴應新添加的變量等時會發生什麼? – James
當我輸入電子郵件和密碼它迴應一個明確的形式,並不能直接到其他功能,我發現mySql數據庫加密爲電子郵件= 0和密碼= 9193ce – user3046785