我试图从我的表回声i现场。但是,当我运行代码下面的错误被表示
消息:试图获得非对象的属性
文件名:视图/ details_view.php
行号:6
我控制器
function index()
{
//using model
$data['list'] = $this->Rfetch_model->getdata();
$this->load->view('Rfetch_view', $data);
}
function get_by_id($id = 0){
$data['info'] = $this->Rfetch_model->getdata_by_id($id);
$this->load->view('details_view', $data);
}
的功能我的模型
function getdata() {
$this->db->select ('subject, id, problem'); // field name
$sql = $this->db->get('info'); // table name
if ($sql->num_rows() >0) {
foreach($sql->result() as $row) {
$data[$row->id] = $row->subject;
}
return $data;
}
}
function getdata_by_id($id = 0){
$this->db->where('id',$id);
$sql = $this->db->get('info');
return $sql->result();
}
}
我的观点(details_view)的功能:
<?php echo $info->problem; ?>
当我打印info.it给出了下面的输出
Array
(
[0] => stdClass Object
(
[id] => 2
[address] => some
[area] => some
[lat] => 1223
[lng] => 2133
[subject] => some
[problem] =>problem
[image] =>
[time] => 2011-08-12 01:09:44
[register_id] => 1
[category_id] => 2
[city_city_id] => 1
[status_status_id] => 0
)
)
这是什么框架? – JJJ
是的。我正在使用codeigneter – brick
你做错了什么?那么,这:http://stackoverflow.com/questions/7039937/echoing-in-the-view-page/7039958 –