为每个实体选择前N个记录

Question

我有一个table像下面 -

---

ID   |  Reported Date                  | Device_ID
-------------------------------------------
1    | 2016-03-09 09:08:32.827         | 1
2    | 2016-03-08 09:08:32.827         | 1
3    | 2016-03-08 09:08:32.827         | 1
4    | 2016-03-10 09:08:32.827         | 2
5    | 2016-03-05 09:08:32.827         | 2

现在,我想要一个top 1 row基于date column每个device_ID

预期产出

---

ID   |  Reported Date                  | Device_ID
-------------------------------------------
1    | 2016-03-09 09:08:32.827         | 1
4    | 2016-03-10 09:08:32.827         | 2

我在用SQL Server 2008 R2.我可以去编写Stored Procedure处理它,但想用简单的查询来做.

****************编辑**************************

回答'Felix Pamittan'运作良好,但'N'只是改为

SELECT
    Id, [Reported Date], Device_ID
FROM (
    SELECT *,
        Rn = ROW_NUMBER() OVER(PARTITION BY Device_ID ORDER BY [ReportedDate] DESC)
    FROM tbl
)t
WHERE Rn >= N

他在评论中提到了这一点,并将其添加到问题中,以便没有人遗漏它.

Answer 1

用途ROW_NUMBER:

SELECT
    Id, [Reported Date], Device_ID
FROM (
    SELECT *,
        Rn = ROW_NUMBER() OVER(PARTITION BY Device_ID ORDER BY [ReportedDate] DESC)
    FROM tbl
)t
WHERE Rn = 1