显示来自页表用户代码点火器的新页表详细信息

Question

我在这个项目中使用CodeIgniter。在我的项目的第一页名为"v_presensi_user_admin_awal"中，我必须显示一个名为"user"的表。数字有"否"列，详细信息按钮有"详细信息"列，用户名有"nama_user"列。当我单击详细信息按钮时，它应该转到"v_presensi_user"以显示我单击的用户的详细信息状态。但它显示出一些错误。

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

以及以下错误用于其他详细信息列。 首先，控制器"Presensi_user"它显示"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);
}

在获取方法"get_pres_user_adm_awal"的模型"presensi_m"中：

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>

如果我单击详细信息按钮，它应该继续控制器"Presensi_user"中称为"详细信息"的方法：

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"v_presensi_user"视图中出现细节：

<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>

请帮助我解决这个问题。我已经尝试并搜索了很多方法，但找不到正确的方法。

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);
}