如何找到oracle中时间戳的区别？

Question

我一定要找到这两个时间戳之间的差异，请帮我

27-FEB-12 02.11.31.910000000 AM, 27-FEB-12 02.11.49.002000000 AM 

Answer 1

减去他们。结果将是一个INTERVAL数据类型，在本例中为17.092秒。

SQL> ed 
Wrote file afiedt.buf 

    1 select to_timestamp('27-FEB-12 02.11.31.910000000 AM', 'DD-MON-RR HH.MI.SS.FF9 AM') - 
    2   to_timestamp('27-FEB-12 02.11.49.002000000 AM', 'DD-MON-RR HH.MI.SS.FF9 AM') 
    3* from dual 
SQL>/

TO_TIMESTAMP('27-FEB-1202.11.31.910000000AM','DD-MON-RRHH.MI.SS.FF9AM')-TO_ 
--------------------------------------------------------------------------- 
-000000000 00:00:17.092000000 

Answer 2

在过去是如何显示你的输出更多的控制，你可以使用EXTRACT：

SQL> SELECT TO_CHAR(EXTRACT(HOUR FROM (x.ts2 - x.ts1)) ,'fm00')      hours 
    2 ,  TO_CHAR(EXTRACT(MINUTE FROM (x.ts2 - x.ts1)) ,'fm00')      minutes 
    3 ,  TO_CHAR(EXTRACT(SECOND FROM (x.ts2 - x.ts1)) ,'fm00.' || RPAD('0',9,'0')) seconds 
    4 FROM (SELECT TO_TIMESTAMP('27-FEB-12 02.11.31.910000000 AM', 'DD-MON-RR HH.MI.SS.FF9 AM') ts1 
    5  ,  TO_TIMESTAMP('27-FEB-12 02.11.49.002000000 AM', 'DD-MON-RR HH.MI.SS.FF9 AM') ts2 
    6  FROM DUAL) x 
    7 ; 

HOU MIN SECONDS 
--- --- ------------- 
00 00 17.092000000 

SQL> 

见http://docs.oracle.com/cd/B28359_01/server.111/b28286/functions052.htm。