怎么使用jQuery解析这段json类型的数据

Question

[{"objName":"datalist","objUser":[{"sUserName":"JSON0","sAddress":"中国-北京-海淀0号"},{"sUserName":"JSON1","sAddress":"中国-北京-海淀1号"},{"sUserName":"JSON2","sAddress":"中国-北京-海淀2号"},{"sUserName":"JSON3","sAddress":"中国-北京-海淀3号"},{"sUserName":"JSON4","sAddress":"中国-北京-海淀4号"}]}]
这是我写的。搞不定啊.
$(function() {
var json = [{"objName":"datalist","objUser":[{"sUserName":"JSON0","sAddress":"中国-北京-海淀0号"},{"sUserName":"JSON1","sAddress":"中国-北京-海淀1号"},{"sUserName":"JSON2","sAddress":"中国-北京-海淀2号"},{"sUserName":"JSON3","sAddress":"中国-北京-海淀3号"},{"sUserName":"JSON4","sAddress":"中国-北京-海淀4号"}]}]
    var data = eval('(' + json + ')');
    alert(data.objName);
    $.each(data.objUser,
    function(i, o) {
        alert(o.sAddress);
    });
})
我要弹出提示：datalist,然后是每个sAddress的值
求高人指点

Answer 1

var json = [{"objName":"datalist","objUser":[{"sUserName":"JSON0","sAddress":"中国-北京-海淀0号"},{"sUserName":"JSON1","sAddress":"中国-北京-海淀1号"},{"sUserName":"JSON2","sAddress":"中国-北京-海淀2号"},{"sUserName":"JSON3","sAddress":"中国-北京-海淀3号"},{"sUserName":"JSON4","sAddress":"中国-北京-海淀4号"}]}];
for(i in json){
     alert(json[i].objName);
     for(s in json[i].objUser){
        alert(json[i].objUser[s].sAddress);
       }
}

Answer 2

json变量本来就是JSON格式的了，LZ为什么还要用var data = eval('(' + json + ')');把这红色这句去了。
$(function() {
    var json = [
                  {
                      "objName":"datalist",
                      "objUser":[
                                  {"sUserName":"JSON0","sAddress":"中国-北京-海淀0号"},
                                  {"sUserName":"JSON1","sAddress":"中国-北京-海淀1号"},
                                  {"sUserName":"JSON2","sAddress":"中国-北京-海淀2号"},
                                  {"sUserName":"JSON3","sAddress":"中国-北京-海淀3号"},
                                  {"sUserName":"JSON4","sAddress":"中国-北京-海淀4号"}
                                ]
                  }
                ];  
    $.each(json[0].objUser,function(i, o) {
        alert(o.sAddress);
    });
})

Answer 3

<script src="http://code.jquery.com/jquery-latest.js"></script>
<script type="text/javascript">
$(function() {
var json = [{"objName":"datalist","objUser":[{"sUserName":"JSON0","sAddress":"中国-北京-海淀0号"},{"sUserName":"JSON1","sAddress":"中国-北京-海淀1号"},{"sUserName":"JSON2","sAddress":"中国-北京-海淀2号"},{"sUserName":"JSON3","sAddress":"中国-北京-海淀3号"},{"sUserName":"JSON4","sAddress":"中国-北京-海淀4号"}]}];
var a=json[0].objUser;
var cnt=json[0].objUser.length;
alert(json[0].objName);
for(var i=0;i<cnt;i++){
alert(a[i].sAddress);
}
})
</script>

Answer 4

楼主你这里不需要eval转换，就楼上说的，你当前json也是object类型了， 不需要转换的，你可以
  console.info(typeof json) 一下 就知道了 ,如果是string,你才需要eval的 
，不过这里个人有点建议不要用eval ,改用new Function的方式
    var b = new Function("return " + json);
    return b.apply(this);
感觉这个兼容性，性能会好点。