Question

我必须在ID列的日期列和分组中选择最新日期和最后一个日期。

这样我就可以获得特定ID的最新日期和最后一个日期。

假设，

-----------------------------
ID          Date
-----------------------------
AA      5/5/2012
AA      6/5/2012
BB      19/5/2012
BB      20/5/2012
BB      18/5/2012
BB      17/5/2012
CC      8/5/2012
CC      19/5/2012
CC      20/5/2012

必需的输出

-----------------------------
ID          Date
-----------------------------
AA      6/5/2012
AA      5/5/2012
BB      20/5/2012
BB      19/5/2012
CC      20/5/2012
CC      19/5/2012

Answer 1

简单使用ROW_NUMBER：

;WITH OrderedRows as (
    SELECT ID,Date,ROW_NUMBER() OVER (PARTITION BY ID ORDER BY Date desc) rn
    from Table
)
select * from OrderedRows where rn <=2

Answer 2

注意：使用UNION ALL INSAL for sql-server-2005。 sql-server-2008 +将使用多值列表。

CREATE TABLE #dates(
    id CHAR(2) NOT NULL,
    dt DATE NOT NULL
);
INSERT INTO #dates(
    id,
    dt
)
SELECT
    'AA','2012-05-05'
UNION ALL
SELECT
    'AA','2012-05-06'
UNION ALL
SELECT
    'BB','2012-05-19'
UNION ALL
SELECT
    'BB','2012-05-20'
UNION ALL
SELECT
    'BB','2012-05-18'
UNION ALL
SELECT
    'BB','2012-05-17'
UNION ALL
SELECT
    'CC','2012-05-08'
UNION ALL
SELECT
    'CC','2012-05-19'
UNION ALL
SELECT
    'CC','2012-05-20';

SELECT
    id,
    dt,
    ROW_NUMBER() OVER (PARTITION BY id ORDER BY dt DESC) AS seq_id
INTO
    #dates_seq
FROM
    #dates;

SELECT
    id,
    dt
FROM 
    #dates_seq
WHERE
    seq_id<=2
ORDER BY
    id,
    dt DESC;

DROP TABLE #dates_seq;
DROP TABLE #dates;

Answer 3

还有一个选择

SELECT *
FROM Table5 t
WHERE t.Date IN(
                SELECT TOP 2 t2.Date
                FROM Table5 t2
                WHERE t.ID = t2.ID
                ORDER BY t2.Date DESC
                )

Answer 4

请检查MSDN solution是否有SQL Server 2005 / SQL Server 2008。

从SQL Server 2005中的日期列中选择最新日期和最后一个日期

4 个答案: