对返回结果进行了统一封装，怎么使用@JsonView注解

Question

 @RequiresPermissions({"manager:view"})
    @RequestMapping("/getManagers")
    @JsonView(Manager.simpleManager.class)
    public Result getManagers(){
        Result result = new Result();
        List<Manager> managers = managerService.selectList(null);
        if(managers!=null){
            result.setCode(1);
            result.setData(managers);
            result.setMsg("查询成功");
        } else {
            result.setCode(0);
            result.setMsg("查询失败");
        }
        return result;
    }

封装的返回

public class Result {
    private String msg;
    private Integer code;
    private Object data;

    public String getMsg() {
        return msg;
    }

    public void setMsg(String msg) {
        this.msg = msg;
    }

    public Integer getCode() {
        return code;
    }

    public void setCode(Integer code) {
        this.code = code;
    }

    public Object getData() {
        return data;
    }

    public void setData(Object data) {
        this.data = data;
    }
}

只有把返回改成List才有效，我这种情况应怎么处理？

Answer 1

     * 查询所有管理员
     * @return
     */

    @RequiresPermissions({"manager:view"})
    @RequestMapping("/getManagers")
//    @JsonView(Manager.simpleManager.class)
    public Result getManagers(){
        Result result = new Result();
        JsonConfig jc=new JsonConfig();
        // 使用setExcludes方法过滤,例如我不想要对象中的roleinfo和departmentInfo：
        jc.setExcludes(new String[]{"password","salt","roles","roleSet","shops","permissionsSet"});
        // 最后转换出来的字符串就会过滤掉我们不要的属性。
        List<Manager> managers = managerService.selectList(null);
        String newArray = JSONArray.fromObject(managers,jc).toString();
        if(managers!=null){
            result.setCode(1);
            result.setData(JSONObject.parse(newArray));
            result.setMsg("查询成功");
        } else {
            result.setCode(0);
            result.setMsg("查询失败");
        }
        return result;
    }

使用JsonConfig过滤解决了