Question

我为此项目使用CodeIgniter。在我的项目“ v_presensi_user_admin_awal”的第一页中，我必须显示一个名为“ user”的表。有“ no”代表数字，“ Detail”代表详细信息按钮，“ nama_user”代表用户名。单击详细信息按钮时，它应转到“ v_presensi_user”以显示我单击的用户的详细信息。但是它显示了一些错误。

A PHP Error was encountered
Severity: Notice

Message: Trying to get property 'nama_user' of non-object

Filename: views/v_presensi_user.php

Line Number: 26

Backtrace:

File: /var/www/html/bolehstation/myreport/application/views/v_presensi_user.php
Line: 26
Function: _error_handler

File: /var/www/html/bolehstation/myreport/application/views/template.php
Line: 183
Function: view

File: /var/www/html/bolehstation/myreport/application/controllers/Presensi_user.php
Line: 33
Function: view

File: /var/www/html/bolehstation/myreport/index.php
Line: 315
Function: require_once

以及其他详细信息的以下错误。首先，控制器“ Presensi_user”显示“ v_presensi_user_admin_awal”：

public function index()
    {
        $data['konten']="v_presensi_user_admin_awal";
        $this->load->model('presensi_m');
        $data['data_presensi']=$this->presensi_m->get_pres_user_adm_awal();
        $this->load->model('status_m');
        $data['data_status']=$this->status_m->get_status();
        $this->load->model('user_m');
        $data['data_user']=$this->user_m->get_user();
        $this->load->model('pengajar_m');
        $data['data_pengajar']=$this->pengajar_m->get_pengajar();
        $this->load->view('template', $data, FALSE);
    }

在模型“ presensi_m”中，该方法获得方法“ get_pres_user_adm_awal”：

public function get_pres_user_adm_awal()
  {
      $data_presensi= $this->db
        ->join('user','user.id_user=presensi.id_user')
        ->join('status','status.id_status=presensi.id_status')
        ->join('pengajar','pengajar.id_pengajar=presensi.id_pengajar')
        ->get('presensi')->result();
      return $data_presensi;
  }

在“ v_presensi_user_admin_awal”中：

<table class="table table-hover table-striped">
   <tr>
      <th>NO</th>
      <th>AKSI</th>
      <th>NAMA SISWA</th>
   </tr>
   <?php
     $no=0;
     foreach ($data_user as $usr) {
       $no++;
       echo '<tr>
         <td>'.$no.'</td>
         <td>'.anchor('Presensi_user/detail/'.$usr->id_user,'<div class="btn btn-info">Detail</div>').'</td>
         <td>'.$usr->nama_user.'</td>
       </tr>';
     }
   ?>
</table>

如果单击详细信息按钮，则它应继续执行控制器“ Presensi_user”中的“详细信息”方法：

public function detail($id_presensi)
    {
        $this->load->model('presensi_m');
        $detail = $this->presensi_m->detail_data($id_presensi);
        $data['detail'] = $detail;
        $data['konten']="v_presensi_user";
        $this->load->view('template', $data, FALSE);
    }

在模型“ presensi_m”中称为方法“ detail_data”的方法：

public function detail_data($id_presensi = NULL)
  {
    $query = $this->db->where('id_presensi', $id_presensi)->get('presensi')->row();
    return $query;
  }

最后，它应该在视图“ v_presensi_user”中显示详细信息：

<table class="table table-hover table-striped">
  <tr>
    <th>NO</th>
    <th>NAMA SISWA</th>
    <th>NAMA PENGAJAR</th>
    <th>HARI</th>
    <th>DATANG</th>
    <th>PULANG</th>
    <th>STATUS</th>
  </tr>
  <?php
    $no=0;
    $no++;
    echo '<tr>
       <td>'.$no.'</td>
       <td>'.$detail->nama_user.'</td>
       <td>'.$detail->nama_pengajar.'</td>
       <td>'.$detail->hari.'</td>
       <td>'.$detail->datang.'</td>
       <td>'.$detail->pulang.'</td>
       <td>'.$detail->status.'</td>
    </tr>'
  ?>
</table>

请帮助我解决此问题。我已经尝试并搜索了许多方法，但是找不到正确的方法。

Answer 1

  public function detail_data($id_presensi = NULL)
  {
    $this->db->select('*');
    $this->db->from('presensi');
    $this->db->where('id_presensi',$id_presensi);
    $query = $this->db->get();
    return $query->result();
  }

    public function detail($id_presensi)
   {
      $this->load->model('presensi_m');
      $data['detail'] = $this->presensi_m->detail_data($id_presensi);
      $data['konten']="v_presensi_user";
      $this->load->view('template', $data);
   }

通过页表用户代码生成器显示新页表详细信息

1 个答案: