Question

我正在尝试创建一个功能，以便用户可以在数据库中更改他们的电子邮件。当我提交表单时，我收到错误，当我尝试提交表单时，登录用户的电子邮件不会更改。

我的控制器：

<?php
class GwController extends CI_Controller {
    // index functie - hierin laad ik de view file 'gegevenswijigen' in het mapje views.
    public function index()
    {
        $this->load->view('gegevenswijzigen');
    }

    function __construct()
    {
         parent::__construct();
         //Laad het Update_model in models folder
         $this->load->model('Update_model');
    }

    //Controller functie om email van een gebruiker te veranderen
     function update_email() 
     {
         $id= $this->input->post('user_id');
         $data = array(
         'email' => $this->input->post('email'));

         $wachtwoord = $_POST['wachtwoord'];

         //check gebruiker in database
         $this->db->select('*');
         $this->db->from('users');
         $this->db->where(array('user_id'=>$_SESSION['user_id'], 'wachtwoord' => $wachtwoord));
         $query = $this->db->get();

         $user = $query->row();
         //Als gebruiker bestaat
        if ($user->user_id) {
            $this->Update_model->update_email($id, $data);
            header ('location:https://kadokado-ferran10.c9users.io/user/profile');
         }
     }
}

我的模特功能：

<?php
class Update_model extends CI_Model{


// Model update functie om email te veranderen
function update_email($id,$data){
$this->db->where('user_id', $id);
$this->db->update('users', $data);
$_SESSION['email'] = $data['email'];
//header ('location:https://kadokado-ferran10.c9users.io/user/profile');
}

}

我的观点：

 <form action="<?php echo base_url() . "GwController/update_email"?>" method="post">
                <tr>
                    <td><h4>E-mail adres wijzigen</h4></td>
                    <td><?php echo form_input(array('type'=>'email','id'=>'email', 'name'=>'email', 'placeholder' => 'Nieuw e-mail adres:', 'size'=>70));?></td>
                    <td><?php echo form_input(array('type'=>'password','id'=>'wachtwoord', 'name'=>'wachtwoord', 'placeholder' => 'Uw wachtwoord:', 'size'=>70));?></td>
                    <td><?php echo form_input(array('type'=>'hidden','id'=>'user_id', 'name'=>'user_id', 'value' => $_SESSION['user_id'], 'size'=>70));?></td>
                    <td><button type="submit" name="emailwijzigen" class="btn btn-primary">Opslaan</button></td>
                </tr>
            </form>

这是我在尝试填写表单并提交时收到的错误：

A PHP Error was encountered
Severity: Notice

Message: Trying to get property of non-object

Filename: controllers/GwController.php

Line Number: 45

Backtrace:

File: /home/ubuntu/workspace/application/controllers/GwController.php
Line: 45
Function: _error_handler

File: /home/ubuntu/workspace/index.php
Line: 315
Function: require_once

这是第45行：

if ($user->user_id) {

感谢任何形式的帮助，谢谢

Answer 1

让我们完成问题排查步骤。

是否设置了$ _SESSION [＆＃39; user_id＆＃39;]？在要检查的函数的开头echo 'sessid:'.print_r($_SESSION['user_id']);。
查询是否会产生任何结果？在echo 'any results?:' . $query->num_rows();

$user = $query->row();

如果没有，请使用echo $this->db->get_compiled_select();而不是$query = $this->db->get();手动检查，并尝试使用phpmyadmin中生成的查询进行手动。
如果phpmyadmin没有返回任何行而num_rows()没有返回任何行，那么条件$this->db->where(array('user_id'=>$_SESSION['user_id'], 'wachtwoord' => $wachtwoord));正在评估数据库中不存在user_id和密码的用户
检查密码字段是否包含echo $_POST['wachtwoord'];

另外请考虑这样的错误情况。您的代码应始终考虑到帐户错误。我已经冒昧地注释你的功能，以便妥善处理错误。

function update_email() {
    $id = $this->input->post('user_id');
    $email = $this->input->post('email');
    $wachtwoord = $this->input->post('wachtwoord');

    // input->post returns null if not set, we need all these variables
    // thus they should all be set
    if (is_null($id) || is_null($email) || is_null($wachtwoord)) {
        exit('Missing parameters'); // handle this nicer with session flash messages
    }

    //check gebruiker in database
    $this->db->select('*');
    $this->db->from('users');
    $this->db->where(array('user_id' => $_SESSION['user_id'], 'wachtwoord' => $wachtwoord));
    $query = $this->db->get();

    // there should only be one valid user for the given
    // user_id and password... otherwise we have no results and $user->user_id
    // will fail or we have a bigger problem on our hands (multiple users w/
    // the same id
    if ($query->num_rows() !== 1) {
        exit('User error');
    }

    $user = $query->row();
    //Als gebruiker bestaat
    // not sure what the point of this conditional statement is...
    // could move inside num_rows()
    if ($user->user_id) {
        $this->Update_model->update_email($id, array('email' => $email));
        header('location:https://kadokado-ferran10.c9users.io/user/profile');
    } else {
        // do something (should always have an else statement in a case like this)
    }
}

请进一步阅读：https://codeinphp.github.io/post/controllers-for-frontend-and-backend-in-codeigniter/

这是处理Admin/LoggedIn和Public部分的更好方法。

更改在CodeIgniter中不起作用的用户数据库信息

1 个答案: