我试图得到像select * from bookdetails where display_id = $id with few foreign key join condition
这样的结果
我已经编写了以下查询,但它显示错误如下:
致命错误:在 line432' 处调用 C:''xampp''htdocs''project 中非对象上的成员函数 num_rows(),即 *if ($query->num_rows()> 0)...
型号.php
public function get_all_book_list_atHomeTop($id, $limit, $start)
{
$this->load->database();
$this->db->limit($limit, $start);
$this->db->get_where('bookdetails', array('display_id' => $id));
//-------join condition ------------------
//------------Ends here Join condition
$query = $this->db->get();
if ($query->num_rows() > 0)
{
foreach ($query->result() as $row)
{
$data[] = $row;
}
return $data;
}
return false;
}
get()
函数中缺少表名:
$query = $this->db->get('bookdetails');
或者你可以简单地用你开头get_where()
语句替换它:
$query = $this->db->get_where('bookdetails',array('display_id'=>$id));