从两个TIMESTAMP中获取时间差(秒)

When you subtract two variables of type TIMESTAMP, you get an INTERVAL DAY TO SECOND which includes a number of milliseconds and/or microseconds depending on the platform. If the database is running on Windows, systimestamp will generally have milliseconds.
If the database is running on Unix, systimestamp will generally have microseconds.

1  select systimestamp - to_timestamp( ‘2012-07-23‘, ‘yyyy-mm-dd‘ )

2*   from dual

SQL> /

SYSTIMESTAMP-TO_TIMESTAMP(‘2012-07-23‘,‘YYYY-MM-DD‘)

---------------------------------------------------------------------------

+000000000 14:51:04.339000000

You can use the EXTRACT function to extract the individual elements of an INTERVAL DAY TO SECOND

SQL> ed

Wrote file afiedt.buf

select extract( day from diff ) days,

extract( hour from diff ) hours,

extract( minute from diff ) minutes,

extract( second from diff ) seconds

from (select systimestamp - to_timestamp( ‘2012-07-23‘, ‘yyyy-mm-dd‘ ) diff

from dual)

SQL> /

DAYS      HOURS    MINUTES    SECONDS

---------- ---------- ---------- ----------

0         14         55     37.936

You can then convert each of those components into milliseconds and add them up

SQL> ed

Wrote file afiedt.buf

1  select extract( day from diff )*24*60*60*1000 +

2         extract( hour from diff )*60*60*1000 +

3         extract( minute from diff )*60*1000 +

4         round(extract( second from diff )*1000) total_milliseconds

5    from (select systimestamp - to_timestamp( ‘2012-07-23‘, ‘yyyy-mm-dd‘ ) diff

6*           from dual)

SQL> /

TOTAL_MILLISECONDS

------------------

53831842

Normally, however, it is more useful to have either the INTERVAL DAY TO SECOND representation or to have separate columns for hours, minutes, seconds, etc. rather than computing the total number of milliseconds between two TIMESTAMP values.

Subtraction between timestamps returns an INTERVAL datatype. You can use the EXTRACT function to return various parts of an interval eg select extract(hour from (timestamp ‘2009-12-31 14:00:00‘ - timestamp ‘2009-12-31 12:15:00‘)) hr from dual; Note: That only
shows the HOUR part, so if the difference is 1 day and 1 hour, this will show 1 not 25. –  Gary Myers Jul 8 ‘09 at 22:42

Another answer:

SQL> @id8

SQL> drop   table holder ;

Table dropped.

SQL> create table holder (

2  beg_date timestamp,

3  end_date timestamp)

4  /

Table created.

SQL> INSERT INTO HOLDER VALUES(to_timestamp(‘2009-07-16:19:00:01.50‘,‘YYYY-MM-DD:HH24:MI:SS.FF‘),

2                        to_timestamp(‘2009-08-17:20:00‘,‘YYYY-MM-DD:HH24:MI‘));

1 row created.

SQL> COMMIT;

Commit complete.

SQL>

SELECT EXTRACT (DAY    FROM (END_DATE-BEG_DATE))*24*60*60+

DELTA

----------

2768398.5

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