我试图通过'id'格式id示例id = GE- dis -001调用数据与Codeigniter的JOIN表,在'where子句'中'未知列' GE '的结果有什么问题?
控制器:
public function detail(){
$id = $this->uri->segment(4);
$detail = $this->mcrud->get_detail($id);
$data = array(
'detail'=>$detail,
'lihat' =>$this->mcrud->lihat_komentar($id),
'isi' =>'instrument/read_views');
$this->load->view('layout/wrapper', $data);
}
模型:
public function get_detail($id){
$sql = "SELECT * FROM tbdetail
JOIN tbkalibrasi ON tbkalibrasi.id = tbdetail.id
JOIN tbsupplier ON tbsupplier.namasupplier = tbdetail.namasupplier
WHERE tbdetail.id = {$id}
";
return $this->db->query($sql)->result_array();
}
似乎$id
变量包含字符串与空格,所以你得到这个错误。改为
public function get_detail($id){
$sql = "SELECT * FROM tbdetail
JOIN tbkalibrasi ON tbkalibrasi.id = tbdetail.id
JOIN tbsupplier ON tbsupplier.namasupplier = tbdetail.namasupplier
WHERE tbdetail.id = '$id'
";
return $this->db->query($sql)->result_array();
}
您的变量id缺少单引号吗?
public function get_detail($id){
$sql = "SELECT * FROM tbdetail
JOIN tbkalibrasi ON tbkalibrasi.id = tbdetail.id
JOIN tbsupplier ON tbsupplier.namasupplier = tbdetail.namasupplier
WHERE tbdetail.id = '{$id}'
";
return $this->db->query($sql)->result_array();
}