我尽量去实现简单的登录应用程序,我得到这个笨错误,但是当我插入的电子邮件和密码登录形式,它以相同的形式作出回应,我无法登录笨简单登录错误
这是我的模型功能:
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" />
您确定密码未加密? – sulmanpucit
会发生什么?你尝试了什么,结果如何?当你回应新添加的变量等时会发生什么? – James
当我输入电子邮件和密码它回应一个明确的形式,并不能直接到其他功能,我发现mySql数据库加密为电子邮件= 0和密码= 9193ce – user3046785