通过AJAX和PHP，提交JQuery Mobile表单(两篇)

发布时间：2020-12-16 01:30:30 所属栏目：百科来源：网络整理

导读：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 h

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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