我在codeigniter中遇到错误,我不知道如何解决这个问题。我的错误是: - 遇到PHP错误
严重性:注意
消息:尝试获取非对象的属性
文件名:views / umparam.php
行号:25
控制器代码:
public function edit($record_mpid)
{
$this->load->model('mparam');
$record = $this->mparam->getAllRecords($record_mpid);
$this->load->view('umparam',['record'=>$record]);
}
}
型号代码:
public function getAllRecords($record_mpid)
{
$query = $this->db->get_where('mparam', array('mpid'=> $record_mpid));
if ( $query->num_rows() > 0){
return $query->row();
}
}
}
查看代码:
<div class="form-group">
<label class="col-md-4 control-label" for="textinput">Parameter</label>
<div class="col-md-6">
<?php echo form_input(['name'=>'parameter','class'=>'form-control','placeholder'=>'Parameter','value'=>set_value('parameter', $record->parameter)]); ?> // line no 25
</div>
<div class="col-md-6">
<?php echo form_error('parameter'); ?>
</div>
</div>
请帮忙。谢谢!!
答案 0 :(得分:0)
在您的控制器功能
上public function edit($record_mpid) {
$this->load->model('mparam');
$data['record'] = $this->mparam->getAllRecords($record_mpid);
$this->load->view('umparam', $data);
}
然后在视图
<div class="form-group">
<label class="col-md-4 control-label" for="textinput">Parameter</label>
<div class="col-md-6">
<?php
$att = array(
'name' => 'field_name',
'class' => 'form-control',
'placeholder' => 'Enter something',
'value' => set_value('field_name', $record->parameter)
);
echo form_input($att);
?>
</div>
<div class="col-md-6">
<?php echo form_error('field_name', '<p>', '</p>'); ?>
</div>
</div>
模型
public function getAllRecords($record_mpid) {
$this->db->where('mpid', $record_mpid);
$query = $this->db->get('mparam');
if ($query->num_rows() > 0) {
return $query->row(); // will return single item;
}
return false;
}