表单提交失败

Question

我有一个表单，我想通过AJAX在div中显示响应。这是来自funds_transfer.php的表单图像。

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我用manual method=post测试它，并用表单提交，它工作得很好。以下是来自funds_transfer_backend.php的部分PHP代码

        $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，但目前它不起作用。下面是我当前无法运行的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获取数据，并将其填充到select box和disabled text box中。这没有问题，它目前工作得很好。

问题是在提交和验证data.response==-1之后，div id="stage中没有响应。我不确定这里有什么问题，可能表单根本没有提交。请提前给予帮助和感谢。

Answer 1

Add id 
<select id="index_num" name="index_num" style="margin-left: 10px;">
because you index_num value is not sending and you get a notice as response with your
json. Also remove from funds_transfer_backend.php this line -> echo "Please complete the form!"

Answer 2

删除此行上的逗号：

amount      : five,

Answer 3

我认为你应该使用$('formname=YourFormName').serialize()；

....
var posted = $('form[name=YourFormName]').serialize();
$.post("funds_transfer_backend.php", posted ,function(data){


if (data.response==-1) { 
    $('#stage').show().html("Please complete the form!"); 
}

    $('#stage').delay(2000).fadeOut();


},"json");
...