我想更新下面的字段,但我的表单不起作用。它不存储数据应该怎么做。我在字段中使用了jquery手风琴,因此它将单击要编辑的项目,然后更新字段然后提交。但它没有用。
查看
foreach($people as $row){
echo "<h3>".$row->service."</h3>";
echo "<form action='".base_url()."some_controller/updateCI' method='post'> <div>Service ID: <input type=text name=id value='".$row->id."' size=27px/><br>Service Name: <input type=text name=name value='".$row->service."'><input type='button' class='classname' value='Save'/></form></div>";
}
&GT?;
CONTROLLER
public function updateCI(){
$this->load->model('some_model');
$id = $this->input->post('id');
$servName = $this->input->post('name');
$success = $this->some_model->updateCI($id,$servName);
if($success == TRUE)
$this->editCI_page(TRUE);
else $this->editCI_page(FALSE);
}
MODEL
public function updateCI($id,$servName){
//$name = $this->db->escape_str($name);
$appID = $this->db->escape_str($id);
$ciName = $this->db->escape_str($servName);
$queryStr = "UPDATE appwarehouse.service SET id='$appID',service='$ciName' WHERE id = '$appID';";
$query = $this->db->query($queryStr);
return $query;
}
答案 0 :(得分:0)
您可以在模型中执行以下操作:
$data = array(
'title' => $title,
'name' => $name,
'date' => $date
);
$this->db->where('id', $id);
$this->db->update('mytable', $data);
我建议你使用可用于codeigniter的活动记录。有关更多信息,请访问以下链接:
http://ellislab.com/codeigniter/user-guide/database/active_record.html