** codeigniter ajax登录问题**

Question

使用ajax和codeigniter登录问题，登录失败，然后在脚本中显示错误消息

ajax脚本不显示任何操作

public function login_autho() {
    $data = array(
        'uname' => $this->input->post('uname'),
        'upassword' => $this->input->post('upassword')
    );
    $result = $this->login_model->login_user($data);
    if ($result == TRUE) {
        //adding data to session 
        $this->load->view('header');
        $this->load->view('user_profile');

    } else if($result==FALSE){
        $this->load->view('user/header');
        $this->load->view('user/login');
   }
}

这是一个用于检查登录和打开profile.php文件的codeigniter控制器

DateTimeZone::listIdentifiers()

Answer 1

这可以解决您的需求

if ($result == TRUE) 
{
    return '1';
} 
else if($result==FALSE)
{
  return '0';
}

在你的ajax succes函数中执行这些

success: function (data) 
{
   if (data == '1')
   {
      window.location.replace("<?php echo base_url(); ?>User_controller/profile");
   }
   else 
   {
       window.location.href("<?php echo base_url(); ?>User_controller/login");
   }
}

Answer 2

从控制器返回值1或0

public function login_autho() 
{
    $data = array(
    'uname' => $this->input->post('uname'),
    'upassword' => $this->input->post('upassword')
    );
    $result = $this->login_model->login_user($data);
    if ($result == TRUE) {
        echo 1;

    } else if($result==FALSE){
        echo 0;
    }
}

Answer 3

谢谢它对我有用

 public function login_autho() 
    {
        $data = array(
        'uname' => $this->input->post('uname'),
        'upassword' => $this->input->post('upassword')
        );
        $result = $this->login_model->login_user($data);
        if ($result == TRUE) {
            echo 1;

    } else if($result==FALSE){
        echo 0;
    }
}