返回錄製最早日期

Question

我使用Aginity Workbench與Netezza數據庫，我試圖與特色三（可維護性）列中的任何一個的IS代碼基於它的最早日期返回的記錄。一個ICS_UID有多個記錄，但我只想返回記錄，其中最早出現的記錄是IS代碼。

下面是我一直在嘗試使用的代碼，但它似乎是返回所有記錄的IS代碼的實例，而不是在where子句中選擇ICS_UID的所有實例。感謝任何幫助或建議。

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

請勿使用GROUP BY。如果你想要一個記錄，則：

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，那麼你可以使用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;