Jquery 子对象的Jqgrid表示
我有一个控制器方法如下Jquery 子对象的Jqgrid表示,jquery,spring-mvc,jqgrid,Jquery,Spring Mvc,Jqgrid,我有一个控制器方法如下 @RequestMapping(value = "/saveUser", method = RequestMethod.POST) @ResponseBody public UserList saveUser(@RequestBody UserList userList) { //some operation } 对象的结构如下所示 public class UserList { private List<User> userList; //
@RequestMapping(value = "/saveUser", method = RequestMethod.POST)
@ResponseBody
public UserList saveUser(@RequestBody UserList userList) {
//some operation
}
对象的结构如下所示
public class UserList {
private List<User> userList;
//getters and setters
}
Public class User{
private Integer id;
private String name;
private Role role;
// getters and setters
}
Public class Role{
private Integer id;
private String name;
private String description;
//getters and setters
}
我如何定义jqgrid,使submit能够命中上述控制器,即我们需要获得上述形式的json。请提及“colModel”,
当我使用像{name:'role.description',index:'role.description',width:55}这样的点运算符时,在用户对象中给出role的空对象
{"userList":[{"id":"1","name":"xyz",
role:{"id":"1","name":"admin","description":"administrator"}}]}