php - 消息:试图在 Codeigniter3 中获取非对象的属性
问题描述
我想显示每个表中的详细数据(作为外键),但不是全部显示。
*我的模型页面中有此代码:
class m_masterinventory extends CI_Model{
public function detail_datasset($id = NULL) {
$query = $this->db->get_where('inv_inventory', array('id' => $id))->result();
return $query;
}
*这是我的视图页面
<table class="table table-hover table-bordered table-striped">
<tr>
<th>ID</th>
<td><?= $asset->initial; ?><?= date('yd', $asset->device); ?>
</td>
</tr>
<tr>
<th>Nama</th>
<td><?= $asset->name; ?></td>
</tr>
<tr>
<th>Organization</th>
<td><?= $asset->organization_id; ?></td>
</tr>
<tr>
<th>Brand</th>
<td><?= $asset->brand; ?></td>
</tr>
<tr>
<th>Category</th>
<td><?= $asset->category; ?></td>
</tr>
<tr>
<th>Model</th>
<td><?= $asset->model; ?></td>
</tr>
<tr>
<th>Serial Number</th>
<td><?= $asset->sn; ?></td>
</tr>
<tr>
<th>Status</th>
<td><?= $asset->status; ?></td>
</tr>
*这是我的控制器页面
public function detail_asset($id){
$this->load->model('m_masterinventory');
$asset = $this->m_masterinventory->detail_datasset($id);
$data['asset'] = $asset;
$this->load->view('inv/detail-master-asset', $data);
}
解决方案
在获取单行时使用row_array()
而不是。result();
$query = $this->db->get_where('inv_inventory', array('id' => $id))->row_array();
并在您看来使用数组 $asset['category']
;代替$asset->category;
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