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Php 用ajax更新数据库_Php_Jquery_Ajax - Fatal编程技术网

Php 用ajax更新数据库

Php 用ajax更新数据库,php,jquery,ajax,Php,Jquery,Ajax,我想用下拉菜单更新我的数据库 我能找到的大多数ajax都是用于退役数据的。有人能帮忙吗 我的php updatestatus.php页面是 include 'includes/session.php'; include 'includes/db_connection.php'; include 'includes/functions.php'; $status = $_POST['status']; $id = $_POST['id']; $sql = "UPDATE orders SET

我想用下拉菜单更新我的数据库 我能找到的大多数ajax都是用于退役数据的。有人能帮忙吗

我的php updatestatus.php页面是

include 'includes/session.php';
include 'includes/db_connection.php';
include 'includes/functions.php';

$status = $_POST['status'];
$id = $_POST['id'];
$sql = "UPDATE orders SET
        status = '$status'
        WHERE id = $id";
我在order.php中的选择框是

<select name="status" id="id" onchange="updateStatus((this.value),<?php echo $row['id']; ?>)">
   <option value="<?php echo $row['status']; ?>"><?php echo $row['status']; ?></option>
   <option value="Order Placed">Order Placed</option>
   <option value="Processing">Processing</option>
   <option value="Dispatched">Dispatched</option>
</select>
行中:

 req.send(data);

在哪里定义
数据
?它必须是包含您的
id
状态的url编码字符串。当我搜索时出现。

数据定义为:

var data = "status="+status+"&id="+id;
还可以将sql更改为:

$sql = "UPDATE orders SET
        status = '".$status."'
        WHERE id =". $id;

为什么要重新发明轮子?使用jQuery。发送请求时是否遇到问题?还是处理请求?
$sql = "UPDATE orders SET
        status = '".$status."'
        WHERE id =". $id;