你应该使用 jQuery 来实现这一点。您可以使用它的ajax()
功能。请访问下面的链接并阅读其完整说明和功能列表,以帮助您解决问题。
这是给您的示例代码:
<hmtl>
<head>
<script type="http://code.jquery.com/jquery-1.5.1.min.js"></script>
</head>
<body>
<div class="my-fake-form">
<input id="posting-value-1" type="text" />
<a id="submit-form-link" href="#submit">Submit this Div!</a>
</div>
</body>
</html>
阿贾克斯代码:
function fake_form_submit ()
{
var post = $('input#posting-value-1').val();
$.ajax({
'url': 'your-php-file.php',
'type': 'POST',
'dataType': 'json',
'data': {post: post},
'success': function(data)
{
if(data.finish)
{
$("div.my-fake-form").attr("innerHTML","Form Submited!");
}
else
{
$("div.my-fake-form").attr("innerHTML","Form Not Submited!");
}
},
beforeSend: function()
{
$(document).ready(function () {
$("div.my-fake-form").attr("innerHTML","Loading....");
});
},
'error': function(data)
{
$(document).ready(function () {
$("div.my-fake-form").attr("innerHTML","ERROR OCCURRED!");
});
}
});
}
$(document).ready(function () {
$('a#submit-form-link').click(function (e) {
e.preventDefault();
fake_form_submit();
});
});
PHP:
<?php
$post = $_POST['post'];
//Do Something with the value!
//On Succes return json encode array!
echo json_encode(array("finish" => true));
?>
AJAX 文档 http://api.jquery.com/jQuery.ajax/
AJAX功能文档 http://api.jquery.com/category/ajax/
AJAX教程 http://www.w3schools.com/Ajax/Default.Asp