jQuery的AJAX表单提交不工作

问题描述：

我想通过AJAX提交使用正常POST提交的表单。的HTML是只是一个标准的形式方法=后和我的jQuery的code是如下：

I'm trying to submit a form through AJAX instead of using the normal POST submission. The HTML is just a standard form with method="post" and my jQuery code is as follows:

jQuery.noConflict();
jQuery(document).ready( function($) {
    var $form = $('#default_contact');

    $form.submit( function() {
        $.ajax({
            type: 'POST',
            url: $form.attr( 'action' ),
            data: $form.serialize(),
            success: function( response ) {
                console.log( response );
            }
        });

        return false;
    });
});

（基于this答案）

我返回false从提交功能，但形式仍然得到提交，我想不出为什么。没有得到任何Firebug的错误。

I'm returning false from the submit function, but the form is still getting submitted and I can't figure out why. Not getting any Firebug errors.

答

您code工作正常，如果HTML是设置正确。这里是我的测试HTML（用PHP），这样你可以比较你自己的。

your code works fine if html is setup right. here's my test html (with php), so you can compare to your own.

<?php
if (!empty($_POST)) {
  die("<pre>" . print_r($_POST, true) . "</pre>");
}
?>
<html>
<head>
  <title>ajax submit test</title>
<script type="text/javascript" src="jquery-1.3.2.min.js"></script>
<script type="text/javascript">
    jQuery.noConflict();
    jQuery(document).ready(function ($) {
        var $form = $('#default_contact');

        $form.submit(function () {
            $.ajax({
                type: 'POST',
                url: $form.attr('action'),
                data: $form.serialize(),
                success: function (response) {
                    alert(response);
                }
            });

            return false;
        });
    });
</script>      
</head>
<body>
  <form id="default_contact" action="formsubmit.php" method="post">
    <input type="hidden" value="123" name="in1" />
    <input type="hidden" value="456" name="in2" />
    <input type="submit" value="send" />
  </form>
</body>
</html>

jQuery的AJAX表单提交不工作

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