1

我想传递对象列表，即具有 userId 和 viewDateTime 字段的用户对象。我们如何生成获取 URL 以从 REST 客户端访问服务？

用户 POJO：

public class User implements Serializable {

    private static final long serialVersionUID = 6407485814769754843L;

    private String userId;

    @JsonProperty(access = Access.READ_ONLY)
    private Date viewDateTime;

    public String getUserId() {
        return userId;
    }

    public void setUserId(String userId) {
        this.userId = userId;
    }

    public Date getViewDateTime() {
        return viewDateTime;
    }

    public void setViewDateTime(Date viewDateTime) {
        this.viewDateTime = viewDateTime;
    }
}

休息控制器：

@RequestMapping(value = APIConstant.GET_ALL, method = RequestMethod.GET)
    public NotificationResponse getAll(@RequestParam("notificationType") NotificationType notificationType,
            AppCodeDTO appCode, @RequestParam(value = "clientDetails", required = false) ClientDetails clientDetails,
            UserListWrapper userlist, @RequestParam(value = "createdByUserId", required = false) String createdByUserId,
            @RequestParam(value = "eventDateTime", required = false) Date eventDateTime,
            PagableSortable pagableSortable) throws QHPlatformException {
        return null;
    }

包装类：

public class UserListWrapper {

    @JsonSerialize(converter=UserConverter.class)
    private List<User> users;

    public List<User> getUsers() {
        return users;
    }

    public void setUsers(List<User> users) {
        this.users = users;
    }

}

我们可以使用 UserWrapper 类传递它，还是可以将 List 作为请求参数传递给控制器​​。请让我知道如何构建 GET 请求的 URL。

谢谢

4

0 回答 0