Return 记录最早日期

Question

我正在使用 Aginity Workbench 和 Netezza 上的数据库，我正在尝试 return 最早日期的记录，其中包含 IS 代码三列（适用性）中的任意一列。一个 ICS_UID 有多个记录，但我只想 return 最早出现的具有 IS 代码的记录。

下面是我一直在尝试使用的代码，但它似乎是 returning 记录具有 IS 代码的所有实例，而不是 ICS_UID 的选择在 where 子句中。感谢任何帮助或建议。

SELECT 
ICS _UID, min(MOVEMENT_DATE) as MOVEMENT_DATE, CURRENT_A_SERVICABILITY_CODE, CURRENT_B_SERVICABILITY_CODE, 
CURRENT_C_SERVICABILITY_CODE 
FROM 
HUB_MOVEMENT
WHERE 
ICS_UID IN (317517607,317962513,etc,etc…)
AND CURRENT_A_SERVICABILITY_CODE = 'IS' OR CURRENT_B_SERVICABILITY_CODE = 'IS' OR CURRENT_C_SERVICABILITY_CODE = 'IS'
GROUP BY 
ICS_UID, CURRENT_A_SERVICABILITY_CODE,
CURRENT_B_SERVICABILITY_CODE,
CURRENT_C_SERVICABILITY_CODE;

Answer 1

不要使用 GROUP BY。如果你想要一条记录，那么：

SELECT m.*
FROM HUB_MOVEMENT m
WHERE ICS_UID IN (317517607,317962513,etc,etc…) AND
      'IS' IN (CURRENT_A_SERVICABILITY_CODE, CURRENT_B_SERVICABILITY_CODE , CURRENT_C_SERVICABILITY_CODE)
ORDER BY MOVEMENT_DATE
LIMIT 1;

如果你想要每个 ICS_UID 一行，那么你可以使用 ROW_NUMBER():

SELECT m.*
FROM (SELECT m.*,
             ROW_NUMBER() OVER (PARTITION BY ICS_UID ORDER BY MOVEMENT_DATE) as seqnum
      FROM HUB_MOVEMENT m
      WHERE ICS_UID IN (317517607,317962513,etc,etc…) AND
            'IS' IN (CURRENT_A_SERVICABILITY_CODE, CURRENT_B_SERVICABILITY_CODE , CURRENT_C_SERVICABILITY_CODE)
     ) m
WHERE seqnum = 1;

Return 记录最早日期

Return Record With Earliest Date

sql

netezza

aginity