DB2:SQL返回一个组中的所有行，这些行在该组的两个最新记录中具有一列的特定值

Question

我有一个DB2表，其中一列(a(的值为PQR或XYZ。

我需要基于C列日期的最新两条记录的值A=PQR的输出。

样品台

A   B     C
--- ----- ----------
PQR Mark  08/08/2019
PQR Mark  08/01/2019
XYZ Mark  07/01/2019
PQR Joe   10/11/2019
XYZ Joe   10/01/2019
PQR Craig 06/06/2019
PQR Craig 06/20/2019

在这个示例表中，我的输出将是Mark和Craig记录的

Answer 1

自11.1起

您可以使用nth_valueOLAP函数。请参阅OLAP规范。

SELECT A, B, C
FROM
(
SELECT 
A, B, C
, NTH_VALUE (A, 1) OVER (PARTITION BY B ORDER BY C DESC ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) C1
, NTH_VALUE (A, 2) OVER (PARTITION BY B ORDER BY C DESC ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) C2
FROM TAB
)
WHERE C1 = 'PQR' AND C2 = 'PQR'

dbfiddle链接。

旧版本

SELECT T.*
FROM TAB T
JOIN 
(
SELECT B
FROM
(
SELECT 
A, B
, ROWNUMBER() OVER (PARTITION BY B ORDER BY C DESC) RN
FROM TAB
)
WHERE RN IN (1, 2)
GROUP BY B
HAVING MIN(A) = MAX(A) AND COUNT(1) = 2 AND MIN(A) = 'PQR'
) G ON G.B = T.B;

Answer 2

是一个简单的解决方案

SELECT A,B,C 
FROM tab
WHERE A = 'PQR'
ORDER BY C DESC FETCH FIRST 2 ROWS only