ajax回调函数接受值问题,求大神们帮忙看看怎么回事

这是输出的data的值:
{"result":
[{"activeCode":"","address":"","area":"??","city":"??","email":"3530341524@qq.com","emailStatus":2,"password":"","phone":"13700000000","pre":"??","sex":1,"time":1233400000000,"updateTime":0,"userCode":"","username":"adminn"},
{"activeCode":"","address":"","area":"??","city":"??","email":"running31335@126.com","emailStatus":0,"password":"","phone":"13700000000","pre":"??","sex":1,"time":0,"updateTime":0,"userCode":"","username":"adminvv"},
{"activeCode":"","address":"","area":"??","city":"??","email":"3530334524@qq.com","emailStatus":0,"password":"","phone":"13700000000","pre":"??","sex":1,"time":0,"updateTime":0,"userCode":"","username":"????"}
]}
$.ajax({
url:"${pageContext.request.contextPath }/manger/manger.do",
type:"post",
beforeSend:function(){
$("#message").text("正在进行查询请稍后...");
return true;
},
success:function(data){
alert(data.result);
/* $.each(data.result,function(index,iteam){
//alert(iteam.username);
}); */
}
});
总是报TypeError: obj is undefined jquery.js (第 583 行,第 4 列)这个异常。怎么解决?

这个事你的前端代码和js产生冲突;可以采用二分法,注释掉一部分代码,排出冲突代码的位置

http://www.cnblogs.com/myjavawork/archive/2011/03/10/1979279.html