JQuery初探---Jquery/Ajax

来源:互联网 发布:张孝祥java面试宝典 编辑:程序博客网 时间:2024/05/16 15:17
<form id="testId" method="post">    <center>     Input ur name Please:   <input name="username" id="username" /><br/>     <a id="isvalid"></a>     Input ur password:   <input name="password" id="password" /><br/>     QQ:   <input name="qq" id="qq" /><br/>     <input type="button" value="submit" onclick="go();" />    </center>  </form>



<script type="text/javascript">function go(){alert("===提交表单===");$.ajax({        cache: true,        type: "POST",        url: 'b.jsp',        data:$('#testId').serialize(),// 你的formid        async: false,        error: function(XMLHttpRequest, textStatus, errorThrown) {        alert(XMLHttpRequest.status);        alert(XMLHttpRequest.readyState);        alert(textStatus);        },        success: function(data) {            //alert("success");            $.messager.confirm('1','2','3');            alert(data);        }    });}</script>


<%@page contentType="text/html;charset=UTF-8"%><%System.out.println("====b.jsp======");System.out.println(request.getParameter("username"));System.out.println(request.getParameter("password"));System.out.println(request.getParameter("qq"));if( request.getParameter("username").equals("hdb") && request.getParameter("password").equals("123456") ){response.getWriter().println("恭喜您登陆成功"); }else{response.getWriter().println("用户名或密码不正确"); }%>


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