DB2：SQL 到 return 组中的所有行在该组的两个最新记录中具有列的特定值

Question

我有一个 DB2 table，其中一个列 (A) 的值是 PQR 或 XYZ。

我需要输出，其中基于列 C 日期的最新两条记录的值为 A = PQR。

样本Table

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

在此示例中 table，我的输出将是 Mark 和 Craig 记录

Answer 1

自 11.1

您可以使用 nth_value OLAP 函数。参考OLAP specification.

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 link.

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

一个简单的解决方案可能是

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

DB2：SQL 到 return 组中的所有行在该组的两个最新记录中具有列的特定值

DB2: SQL to return all rows in a group having a particular value of a column in two latest records of this group

db2

db2-luw

自 11.1

旧版本