使用AJAX/PHP/MySQL批准/拒绝记录

我有一个表格格式的信息列表。现在我在这个表中添加了一列，它有两个选项approve&reject的下拉列表。现在，当他从下拉列表中选择选项时，我想更新单个记录的状态，我想使用AJAX进行更新

PHP代码：

<form name="post_action" method="post" action="" id="post_action">
<select name="post_status" id="<?=$row['id']; ?>" style="width:175px; float:left; height:25px;" >
<option selected="selected" disabled="disabled">--Select--</option>
<option value="1">Approved</option>
<option value="2">Rejected</option>
<option value="3">Pending</option>
</select>

选择如下：

<select name="post_status" id="<?=$row['id']; ?>" onchange="change_status(<?=$row['id']; ?>)" style="width:175px; float:left; height:25px;" >
<option selected="selected" disabled="disabled">--Select--</option>
<option value="1">Approved</option>
<option value="2">Rejected</option>
<option value="3">Pending</option>
</select>

JQuery

var serializedData = $form.serialize();
$("input[type=select]").change(function() {
    $.ajax({//Submit the value of the drop_down
        url: "/updateDB.php",
        type: "post",
        data: serializedData
        }).done(function() {
        alert('Updated!');
    });
});

PHP

javascript/jquery必须触发
onchange
事件，然后使用AJAX将其发送到PHP控制器/文件以更新数据库记录。
var serializedData = $form.serialize();
$("input[type=select]").change(function() {
    $.ajax({//Submit the value of the drop_down
        url: "/updateDB.php",
        type: "post",
        data: serializedData
        }).done(function() {
        alert('Updated!');
    });
});

<?php
     if(isset($_POST['id']){
        $id = $_POST['id'];
        $post_status = $_POST['post_status'];
        $sql = "UPDATE mytable SET myvalue = '$post_status' WHERE id = $id";
        //Connect to database
        $link = mysqli_connect("myhost","myuser","mypassw","mydb") or die("Error " . mysqli_error($link));
        //Execute the Query
        $result = $link->query($sql);
        //Using the $result object you can determine if the update was successful or not and alert the user of success/failure

?>