这篇文章主要讲解了“如何解决ajax返回验证的时候总是弹出error错误的问题”,文中的讲解内容简单清晰,易于学习与理解,下面请大家跟着小编的思路慢慢深入,一起来研究和学习“如何解决ajax返回验证的时候总是弹出error错误的问题”吧!
发一个简单案例:
前台:
<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%> <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> <html> <head> <title>用户登录</title> <script type="text/javascript" src="../js/jquery-easyui-1.3.5/jquery.min.js"></script> <script type="text/javascript" src="../js/jquery-easyui-1.3.5/jquery.easyui.min.js"></script> <link rel="stylesheet" href="../js/jquery-easyui-1.3.5/themes/default/easyui.css" type="text/css"></link> <link rel="stylesheet" href="../js/jquery-easyui-1.3.5/themes/icon.css" type="text/css"></link> <script type="text/javascript" src="../js/jquery-easyui-1.3.5/locale/easyui-lang-zh_CN.js"></script> <meta http-equiv="content-type" content="text/html;charset=UTF-8" /> <script type = "text/javascript" charset = "UTF-8"> $(function(){ var loginDialog; loginDialog = $('#loginDialog').dialog({ closable : false , // 组件添加属性:让关闭按钮消失 //modal : true, //模式化窗口 buttons : [{ text:'注册', handler:function(){ } }, { text:'登录', handler:function(){ $.ajax({ url:'../servlet/Login_Do', data :{ name:$('#loginForm input[name=name]').val(), password:$('#loginForm input[name=password]').val() }, dataType:'json', success:function(r){ //var dataObj=eval("("+data+")"); alert("进来了"); }, error:function(){ alert("失败"); } }); //alert(data) } }] }); }); </script> </head> <body style=”width:100%;height:100%;" > <div id = "loginDialog" title = "用户登录" style = "width:250px;height:250px;" > <form id = "loginForm" method = "post"> <table> <tr> <th>用户名 :</th> <td><input type = "text" class = "easyui-validatebox" data-options="required:true" name = "name"><br></td> </tr> <tr> <th>密码: </th> <td> <input type = "password" class = "easyui-validatebox" data-options="required:true" name = "password"><br></td></td> </tr> </table> </form> </div> </body> </html>
后台:
public class Login_Do extends HttpServlet { public void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { this.doPost(request, response); } public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { request.setCharacterEncoding("UTF-8"); response.setCharacterEncoding("UTF-8"); String name =request.getParameter("name"); String password = request.getParameter("password"); String js = "{\"name\":name,\"password\":password}"; PrintWriter out = response.getWriter(); JSONObject json = new JSONObject(); json.put("name",name); out.print(json.toString()); response.getWriter().write(json.toString()); } }
点击登录时:
解决办法:弹出error信息一般有两种可能:
第一种:url错误,后台直接得不到值
可以用火狐的firebug查看:如果响应了信息,则不是这个问题,那么就有可能是第二种情况:
返回数据类型错误:
在我这个例子中,返回的数据无意中打印了两次,这两句删去一句就好了:
out.print(json.toString()); response.getWriter().write(json.toString());
造成了错误。这时在firebug显示的信息是:
感谢各位的阅读,以上就是“如何解决ajax返回验证的时候总是弹出error错误的问题”的内容了,经过本文的学习后,相信大家对如何解决ajax返回验证的时候总是弹出error错误的问题这一问题有了更深刻的体会,具体使用情况还需要大家实践验证。这里是亿速云,小编将为大家推送更多相关知识点的文章,欢迎关注!
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