datagrid控件展示JSON嵌套对像

{"total":2,"rows":["value":0,"department":{"depDesc":"技术部","depId":10,"depLevel":2,"depName":"产品与技术部","parentId":0,"path":"0.10.","version":null},"email":"admin@panguso.com","empId":5,"enabled":true,"familyName":"超级管理员","firstKeyColumnName":"empId"}]}

如上面JSON对像,我现在要在JQuery的datagrid控件,显示department对像下面的depDesc属性值,应该怎样写代码呢,通过field:'department.depDesc' 都显示不了,直接填写field:'department'显示出来的是[Object Object]

好像可以department["depDesc"]这样写。
你可以看看javascript访问对象的方式有哪几种。。

'rows.department.depDesc' 就可以了，你试试

JsonObject.....完全可以满足你

补充下，你返回都是[Object Object]了，下一步 for(var obj in Object){
Object.obj
}
就可以了撒

这样写好点
{ field: 'depDesc', title: '部门', width: 120,
    formatter: function(val, rec) {
        return rec.department == null ? "" : rec.department.depDesc;
    }
}