getting UNIX epoch time

[Danny Fang]

Hi,

I'm attempting to obtain the UNIX epoch time(since 1970) on a HPUX machine. I've tried with the 'truss' utility i.e.
/usr/bin/truss /usr/bin/date 2>&1|/usr/bin/awk '/time/ {print $NF}'
1159448101

On a GNU platform, I could do:
mz[t1]$ date '+%s'
1159448106

Unfortunately, I do have the GNU date/utilities installed on my HPUX-11 machine.

May I know what is the more effecient or simple way to obtain the UNIX epoch time, instead of using the truss?

Thanks
Danny

[A. Clay Stephenson]

perl -e 'print time()'

[James R. Ferguson]

Hi Danny:

use Perl;

# perl -le 'print time'

# perl -le 'print scalar localtime(time)'

# perl -le 'print scalar localtime(86400)'

Regards!

...JRF...

[Thomas Elsäßer]

echo "0d995462197=Y" | adb
2001 Jul 18 15:16:37

[Danny Fang]

Hi Thomas,

With regards to your solution posted below,
echo "0d995462197=Y" | adb
2001 Jul 18 15:16:37

could you explain what does the "0d995462197=Y" value mean when passed to the debugger?

Thanks
Danny

[James R. Ferguson]

Hi Danny:

Passing "0d995462197=Y" to the debugger ('adb') says to take the *decimal* (0d) number (995462197) and convert it to a time_t object. This is documented in the 'adb' manpages:

http://www.docs.hp.com/en/B2355-60105/adb.1.html

Regards!

...JRF...