我创建了一个类User，可以这么取到数据
function initUser(){
var code = $("#code").val();
var name = $("#name").val();
$("#list").datagrid({
url:"/userWeb/getUserList",
queryParams :{
code : code,
name : name
},
singleSelect: false,
columns:[[
{field:"id",checkbox:"true"},
{field:"code",align:'center',title:"手机号码",width:200},
{field:"name",align:'center',title:"昵称",width:200},
{field:"age",align:'center',title:"年龄",width:200}
]],
//分页工具
pagination:true,
//工具条件
toolbar:"#toolbar"
});
如果我的user中还有一个addressinfo节点，那我怎么展示addresssinfo下的address
@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
private Long id;
private String code;
private String name;
private Integer age;
@Column(name = "system_time")
private Date systemTime;
@Transient
private AddressInfo addressInfo;
public class AddressInfo {
private String address;
public String getAddress() {
return address;
}
public void setAddress(String address) {
this.address = address;
}
求大神指点

各位大神，能不能指点一二

获得的json是这样的
{
"total": 22,
"rows": [
{
"id": 9,
"code": "9",
"name": "BB",
"age": 11,
"systemTime": null,
"addressInfo": {
"address": "这是中国西北部的城市"
}
}
]
}

自问自答。。。
百度了，有两种方式，一种是该表easyui源码，第二种是做转换
function initUser(){
var code = $("#code").val();
var name = $("#name").val();
$("#list").datagrid({
url:"/userWeb/getUserList",
queryParams :{
code : code,
name : name
},
data :[
],
columns:[[
{field:"id",checkbox:"true"},
{field:"code",align:'center',title:"手机号码",width:200},
{field:"name",align:'center',title:"昵称",width:200},
{field:"age",align:'center',title:"年龄",width:200},
{field:"addressInfo.address",align:'center',title:"地址",width:200,
formatter:function(value,row,index){
if(row.addressInfo){
return row.addressInfo.address;
}
return "";
}}
]],
singleSelect: false,
//分页工具
pagination:true,
//工具条件
toolbar:"#toolbar"
});
}