解決ajax返回驗證的時候總是彈出error錯誤的方法
編輯：AJAX基礎知識
發一個簡單案例：
前台：
<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>用戶登錄</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 , // 組件添加屬性：讓關閉按鈕消失
//modal : true, //模式化窗口
buttons : [{
text:'注冊',
handler:function(){
}
},
{
text:'登錄',
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("進來了");
},
error:function(){
alert("失敗");
}
});
//alert(data)
}
}]
});
});
</script>
</head>
<body style=”width:100%;height:100%;" >
<div id = "loginDialog" title = "用戶登錄" style = "width:250px;height:250px;" >
<form id = "loginForm" method = "post">
<table>
<tr>
<th>用戶名 :</th>
<td><input type = "text" class = "easyui-validatebox" data-options="required:true" name = "name"><br></td>
</tr>
<tr>
<th>密碼: </th>
<td> <input type = "password" class = "easyui-validatebox" data-options="required:true" name = "password"><br></td></td>
</tr>
</table>
</form>
</div>
</body>
</html>
後台：
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());
}
}
點擊登錄時：
解決辦法：彈出error信息一般有兩種可能：
第一種：url錯誤，後台直接得不到值
可以用火狐的firebug查看：如果響應了信息，則不是這個問題，那麼就有可能是第二種情況：
返回數據類型錯誤：
在我這個例子中，返回的數據無意中打印了兩次,這兩句刪去一句就好了：
out.print(json.toString()); response.getWriter().write(json.toString());
造成了錯誤。這時在firebug顯示的信息是：
以上就是為大家分析的用ajax返回驗證的時候總是彈出error的原因，希望對大家解決此類問題有所幫助。
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