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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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