解决ajax返回验证的时候总是弹出error错误的方法-前端开发

这篇文章主要介绍了解决ajax返回验证的时候总是弹出error错误的方法，感兴趣的小伙伴们可以参考一下

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="Li4vanMvanF1ZXJ5LWVhc3l1aS0xLjMuNS9qcXVlcnkubWluLmpz"></script> 
   <script type="text/javascript" src="Li4vanMvanF1ZXJ5LWVhc3l1aS0xLjMuNS9qcXVlcnkuZWFzeXVpLm1pbi5qcw=="></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="Li4vanMvanF1ZXJ5LWVhc3l1aS0xLjMuNS9sb2NhbGUvZWFzeXVpLWxhbmctemhfQ04uanM="></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 = "55So5oi355m75b2V" 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：


out.print(json.toString()); 
response.getWriter().write(json.toString());

造成了错误。这时在firebug显示的信息是：

以上就是为大家分析的用ajax返回验证的时候总是弹出error的原因，希望对大家解决此类问题有所帮助。