Oracle SQL，格林威治标准时间 (UTC) 中的日期和时间

Question

This coding is giving the correct answer for SYSDATE

select round((SYSDATE - date '1970-01-01')*24*60*60) from dual;

1662482430 (seconds) 

I need to return the date / time in GMT/UTC.

I need help with the syntax.

Thanks, Pete

Answer 1

If you use SYSTIMESTAMP instead of SYSDATE then that value will be in your DB time zone, and you can then convert that to UTC with at time zone :

SYSTIMESTAMP at time zone 'UTC'

and cast that back to a date:

cast(SYSTIMESTAMP at time zone 'UTC' as date)

and then use that in your calculation:

select round((cast(SYSTIMESTAMP at time zone 'UTC' as date) - date '1970-01-01')*24*60*60)
from dual;

db<>fiddle