前端之家收集整理的这篇文章主要介绍了
通过AJAX和PHP，提交JQuery Mobile表单(两篇)，
前端之家小编觉得挺不错的，现在分享给大家，也给大家做个参考。
File name: callajax.PHP
[PHP] view plaincopy
<?PHP
$firstName = $_POST[firstName];
$lastName = $_POST[lastName];
echo("First Name: " . $firstName . " Last Name: " . $lastName);
?>
File name: index.PHP
[html] view plaincopy
<!DOCTYPE html>
<html>
<head>
<title>Submit a form via AJAX</title>
<link rel="stylesheet" href="http://code.jquery.com/mobile/1.0a4/jquery.mobile-1.0a4.min.css" />
<script src="http://code.jquery.com/jquery-1.5.2.min.js"></script>
<script src="http://code.jquery.com/mobile/1.0a4/jquery.mobile-1.0a4.min.js"></script>
</head>
<body>
<script>
function onSuccess(data,status)
{
data = $.trim(data);
$("#notification").text(data);
}
function onError(data,status)
{
// handle an error
}
$(document).ready(function() {
$("#submit").click(function(){
var formData = $("#callAjaxForm").serialize();
$.ajax({
type: "POST",url: "callajax.PHP",cache: false,data: formData,success: onSuccess,error: onError
});
return false;
});
});
</script>
<!-- call ajax page -->
<div data-role="page" id="callAjaxPage">
<div data-role="header">
<h1>Call Ajax</h1>
</div>
<div data-role="content">
<form id="callAjaxForm">
<div data-role="fieldcontain">
<label for="firstName">First Name</label>
<input type="text" name="firstName" id="firstName" value="" />
<label for="lastName">Last Name</label>
<input type="text" name="lastName" id="lastName" value="" />
<h3 id="notification"></h3>
<button data-theme="b" id="submit" type="submit">Submit</button>
</div>
</form>
</div>
<div data-role="footer">
<h1>GiantFlyingSaucer</h1>
</div>
</div>
</body>
</html>
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
jQuery mobile 表单提交
http://blog.csdn.net/tjpu_lin/article/details/28394253
最近在做手机页面用jQuery mobile开发，在用到form表单提交到时遇到了问题。
后台是用servlet进行处理的，想通过Servlet里面的重定向进行页面跳转发现在手机上根本没有用，出现errorpage提示信息。
查询网上资料以及jQuery mobile API得知 jQuery mobile表单提交默认是ajax提交，所以页面跳转写在servlet里面根本就不会实现页面跳转功能。
于是我按照教程在form里面加了 data-ajax=“false”这一属性，发现别说是页面跳转不行，就连后台的数据库操作都做不了，报了500错误。
想了好久既然用ajax提交，那么就用ajax进行页面跳转
[html] view plaincopy在CODE上查看代码片派生到我的代码片
<script type="text/javascript">
$(document).ready(function () {
$("#submitbtn").click(function(){
cache: false,$.ajax({
type: "POST",url: "Feedback",data: $('#Feedbackform').serialize(),success:function(data){
$.mobile.changePage("success.html");
}
});
});
});
</script>
[html] view plaincopy在CODE上查看代码片派生到我的代码片
<form method="post" id="Feedbackform">
t;span style="white-space:pre"> </span>//相关表单元素
<input type="button" id="submitbtn" value="提交">
</form>
注意的是js里面的data
[html] view plaincopy在CODE上查看代码片派生到我的代码片
$('#Feedbackform').serialize()
是必须要有的，不然servlet里面用requset.getParameter接受的数据是null，ajax和后台成功交互后用changePage跳转到成功后显示的页面。
</pre><pre code_snippet_id="1849295" snippet_file_name="blog_20160824_1_2775332" name="code" class="html">
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