Question

我尝试使用user_id显示其他用户信息（例如姓名和电子邮件），从而建立到视图页面的链接。这是我试图制作的链接：

<?php foreach($userdetail_list as $row) { ?>
    <tr>
        <td><a href="<?php echo base_url() . 'User/profiel_user/'.$row['user_id']?>">
    </tr>
<?php } ?>

现在我正在尝试收集属于我点击的user_id的用户信息。当我点击链接时，我确实看到该链接更改为正确的用户：User / profiel_user / 6 但它说404找不到页面

我的User_model文件：

<?php
class User_model extends CI_Model {
    public function getUserInfoByEmail($email) {
        $q = $this->db->get_where('users', array('email' => $email), 1);  
        if($this->db->affected_rows() > 0){
            $row = $q->row();
            return $row;
        }else{
            error_log('no user found getUserInfo('.$email.')');
            return false;
        }
    }
    public function getUserInfo($user_id) {
        $q = $this->db->get_where('users', array('user_id' => $user_id), 1);  
        if($this->db->affected_rows() > 0){
            $row = $q->row();
            return $row;
        }else{
            error_log('no user found getUserInfo('.$user_id.')');
            return false;
        }
    }
    public function getdata() {
        $this->db->where('user_id', $id);
        $this->db->from('users');
        $query = $this->db->get();
        if($query->num_rows()>0)
        {
           return $query->result_array();
        }
    }
}
?>

我的完整User.php控制器文件：

<?php
    class User extends CI_Controller {

    public function index() {
        $this->load->view('profile', $data);
    }
    public function __construct() {
        parent::__construct();
        if ($_SESSION['user_logged'] == FALSE) {
            $this->session->set_flashdata("error", "Please login first to view this page!! ");
            redirect("auth/login");
        }
    }
    public function userdetails($user_id) {
        //load the User_model
        $this->load->model('User_model');

        //call function getdata in de Product_model
        $data['userdata_list'] = $this->User_model->getdata();

        //get product details
        $data['user'] = $this->User_model->get_user_info($user_id);

        //laad view
        $data['main_content'] = 'profiel_user';
        $this->load->view('profiel_user', $data);
    }
    public function profile() {

        $this->load->model('User_model');
        if ($_SESSION['user_logged'] == FALSE) {
            $this->session->set_flashdata("error", "Please login first to view this page!! ");
            redirect("auth/login");
        }

        $this->load->view('profile');
    }
}
?>

链接链接到的视图页面（profiel_user.php）上有什么：

<?php include_once ('templates/header.php'); ?>
    <div class="container-fluid">
        <div class="row">
            <div class="col-lg-12 bg-warning" style="font-size:25px">
                <center>Gebruikersprofiel</center>
            </div>
        </div>
    </div>
<?php foreach($userdata_list as $row) { ?>
    <tr>
        <td><?php echo $row['email']; ?></td>
     </tr>
<?php } ?>
<?php include_once ('templates/header.php'); ?>
<div class="container-fluid">
    <div class="row">
        <div class="col-lg-12 bg-warning" style="font-size:25px">
            <center>Gebruikersprofiel</center>
        </div>
    </div>
</div>

我希望有人能弄清楚我做错了什么以及为什么我不能使用user_id来回显或显示其他user_information。

数据库：表名：用户而user_id是我的主键

Answer 1

您的方法名称为userdetails而不是profiel_user，因此请更改您的广告

<td><a href="<?php echo base_url() . 'User/profiel_user/'.$row['user_id']?>">

到

<td><a href="<?php echo base_url() . 'User/userdetails/'.$row['user_id']?>">

使用user_id在codeigniter

1 个答案: