2013-11-25 31 views
-1

似乎無法弄清楚這一點。只需要2個oracle時間戳列之間的分鐘差異即可。如何從2個時間戳列中獲得分鐘數的差異?

END_TS         START_TS 
--------------------------------------------------------------------------- --------------------------------------------------------------------------- 
11-NOV-13 01.52.14.000000 PM       11-NOV-13 01.51.14.000000 PM 
11-NOV-13 02.20.47.000000 PM       11-NOV-13 02.19.47.000000 PM 
18-NOV-13 12.44.54.000000 PM       18-NOV-13 12.34.02.000000 PM 
22-NOV-13 12.02.09.000000 AM       22-NOV-13 12.02.08.000000 AM 

回答

0
SELECT ROUND((END_TS - START_TS) * 24 * 60, 0) DifferenceInMinutes 
FROM YourTable 
+0

我得到的結果的形式: DIFFERENCEINMINUTES ----------------- -------------------------------------------------- -------- +000000000 00:00:00.000000000 +000000000 00:00:00.000000000 +000000010 20:48:00.000000000 +000000000 00:24:00.000000000 +000000000 00:00:00.000000000 – KingKongFrog

+0

增加了ROUND()以在十進制之後將分鐘舍入爲0。現在試試。 – Shiva

0
SELECT (CAST(END_TS AS DATE) - CAST(START_TS AS DATE)) * 24 * 60 from dual 

和圓形它如果需要

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