用struts2和jquery.ajax进行异步请求的时候在谷歌浏览器的console出现了
Failed to load resource: the server responded with a status of 500 (Internal Server Error)而struts2中action中的方法正常执行并且能查询到数据到时这些数据不能返回到页面上,即jquery.ajax中的回调函数没有执行
请问该怎么解决[b][/b]
struts2中的action中的方法如下:
public String quer(){
System.out.println("进了");
message="zen";
propertylist = propertyservice.querypropertyall(pu);
return SUCCESS;
}
struts.xml中的配置如下:
<package name="ddd" extends="json-default">
<action name="json" class="propertyAction" method="quer">
<result type="json"></result>
</action>
</package>
jsp页面的类容如下:
<%@ page language="java" import="java.util.*" pageEncoding="ISO-8859-1"%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<base href="<%=basePath%>">
<title>My JSP 'Jquerytest.jsp' starting page</title>
<meta http-equiv="pragma" content="no-cache">
<meta http-equiv="cache-control" content="no-cache">
<meta http-equiv="expires" content="0">
<meta http-equiv="keywords" content="keyword1,keyword2,keyword3">
<meta http-equiv="description" content="This is my page">
<!--
<link rel="stylesheet" type="text/css" href="styles.css">
-->
<script type="text/javascript" src="js/Jquery-easyui-1.2.6/jquery-1.7.2.min.js"></script>
<script type="text/javascript">
function a(){
alert("!!!!!!!!!!!!!!");
$.ajax({
url:'json',
dataType:'json',
data:'pu.currentPage=1',
success:update
});
function update(){
alert("~~~~~~~~~~~~~~~~~~~~");
}
}
</script>
</head>
<body>
<input type="button" value="ddd" onclick="a()">
</body>
</html>
图片是我所用的json的jar包
我的解决办法是:在action中先将该list转换成json字符串,然后再在前台页面接收,然后在转换成oject对象
即action中的添加JSONArray jsonArray = JSONArray.fromObject(articleList);message = jsonArray.toString();
script中添加:
function update(data){
var articletitle= eval('(' + data + ')');
}
- 大小: 7.4 KB
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