返回录制最早日期

Question

我使用Aginity Workbench与Netezza数据库，我试图与特色三（可维护性）列中的任何一个的IS代码基于它的最早日期返回的记录。一个ICS_UID有多个记录，但我只想返回记录，其中最早出现的记录是IS代码。

下面是我一直在尝试使用的代码，但它似乎是返回所有记录的IS代码的实例，而不是在where子句中选择ICS_UID的所有实例。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

请勿使用GROUP BY。如果你想要一个记录，则：

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; 

如果你想每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;