我有一个表单,我想div通过 AJAX 显示响应。这是来自 的表格图像funds_transfer.php。

我用手册测试它method=post并用表单提交,它工作得很好。这是来自的部分 PHP 代码funds_transfer_backend.php:
        $index_num = $_POST['index_num'];
        $to_account_num = $_POST['recipient'];
        $amount = $_POST['amount'];
        if ($amount == '' || $to_account_num == '' || $index_num == -1){
            echo "Please complete the form!";
            $response = -1;
        }
        else {  
                    // other code goes here..
        $display = array('response' => $response); // for ajax response later
        echo json_encode($display);
PHP给了我这个输出:
Please complete the form!{"response":-1}
现在我想用 AJAX 实现它,但它目前不起作用。这是我目前没有工作的 html + jQuery 代码:
<script type="text/javascript" src="http://code.jquery.com/jquery-1.9.0.min.js"></script>
<script type="text/javascript">
 function update() {
      var    two = $('#index_num').val();
      var    three = $('#recipient_box').val();
      var    five = $('#amount_box').val();
 $.post("funds_transfer_backend.php", { 
    index_num   : two,
    recipient   : three, 
    amount      : five
},function(data){
    if (data.response==-1) { 
        $('#stage').show().html("Please complete the form!"); 
    }
        $('#stage').delay(2000).fadeOut();
},"json");
        } 
</script>
//other code goes here..
<p>Transfer your funds to other account
    <script type="text/javascript">
    // Pre populated array of data
    var myData = new Array();
    <?php
                $sql="SELECT * FROM `account` WHERE client_id='$id'";
                $result=mysqli_query($conn, $sql);
                while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { 
                        echo "myData.push('".$row['funds']."');";
                        $result_array[] = $row['id'];
                } 
    ?>
</script>
<form id="example" name="example">
        Select your account<select id="selector" name="index_num" style="margin-left: 10px;">
            <option value="-1">Account Number</option>
    <?php //echo $result_array[0]; 
    $num = count($result_array) - 1;
    $i=0;
    while($i<=$num)
      {
      echo "<option value=\"$i\">$result_array[$i]</option>";
      $i++;
      }
    ?>
        </select>
        <br />
       Funds : RM <input type="text" id="populateme" name="funds" disabled/><br>
       Recipient Account Number <input type="text" id="recipient_box" name="recipient" />  <br> 
       Amount : RM <input type="text" id="amount_box" name="amount"/><br>    
    <input type="button" value="Submit" onclick="update();">
    <input type="reset" value="Reset">
    </form>
    <div id="stage" style="background-color:#FF6666; padding-left:20px; color: white;"></div> 
    <script type="text/javascript">
        document.example.selector.onchange = updateText;
        function updateText() {
          var obj_sel = document.example.selector;
          document.example.populateme.value = myData[obj_sel.value];
    }
    </script>
</p>
上面的 SQL 查询将从 db 中获取数据并将其填充到select boxand中disabled text box。没问题,它目前运行良好。
div id="stage问题是提交和验证后没有响应data.response==-1。我不确定这里有什么问题,可能表格根本没有提交。请提前帮助和感谢。