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How to solve the problem that error always pops up when ajax returns verification

亚连
Release: 2018-05-24 11:29:28
Original
2573 people have browsed it

This article mainly introduces the method to solve the problem that error always pops up when ajax returns verification. Interested friends can refer to it

Send a simple case:
Front desk:

<%@ 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 = $(&#39;#loginDialog&#39;).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:&#39;注册&#39;, 
         handler:function(){ 
            
         } 
       }, 
       { 
         text:&#39;登录&#39;, 
         handler:function(){ 
            $.ajax({ 
             url:&#39;../servlet/Login_Do&#39;, 
             data :{ 
                name:$(&#39;#loginForm input[name=name]&#39;).val(), 
                password:$(&#39;#loginForm input[name=password]&#39;).val() 
               }, 
             dataType:&#39;json&#39;, 
             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%;" > 
    <p 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>  
    </p> 
 </body> 
</html>
Copy after login

Backend:

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()); 
  } 
}
Copy after login

When you click to log in:

Solution: There are generally two possibilities for pop-up error messages :
The first type: url error, the value cannot be obtained directly in the background
You can use Firefox's firebug to check: If the message is responded to, this is not the problem, then it may be Second case:
Return data type error:
In my example, the returned data was accidentally printed twice. Just delete one of these two sentences:

out.print(json.toString()); 
response.getWriter().write(json.toString());
Copy after login

caused an error. The information displayed in firebug at this time is:

The above is what I compiled for everyone. I hope it will be helpful to everyone in the future.

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