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Operating Systems AIX time convert
# 8  
Old 05-21-2008
Quote:
Originally Posted by b_manu78
If i give perl -le 'print scalar localtime(1211381313);' , then the output is correct...Wed May 21 11:18:33 2008.
Are you sure that it's the correct date? When I convert it my way, it give me:

GNU Awk 3.1.5
$ TZ=UTC awk 'BEGIN {print strftime("%Y-%m-%d %T", 1211381313)}'
> 2008-05-21 14:48:33

date (GNU coreutils) 5.97
$ date --utc --date "1970-01-01 1211381313 sec" "+%Y-%m-%d %T"
> 2008-05-21 14:48:33
# 9  
Old 05-21-2008
You are not passing $lastlog1 to Perl correctly, so it's effectively doing localtime(0). (The date it prints is Jan 1 1970 UTC 00:00:00 converted to your local time zone.)

Code:
perl -le 'print scalar localtime(shift)' $lastlog1

If you are using awk anyway, the awk solution ripat posted would seem ideal.

Code:
lsuser -a time_last_login root|awk -F= '{print strftime("%Y-%m-%d %T",$2)}'

# 10  
Old 05-21-2008
Just for the record: the problem is here:

Quote:
Originally Posted by b_manu78
perl -le 'print scalar localtime($lastlog1);'
You try to expand variable ($lastlog1) inside single quotes. To prevent special characters like "$" from being interpreted by the shell is exactly what single quotes have been invented for.

The line will probably work writing it that way:

Code:
perl -le 'print scalar localtime('"$lastlog1"');'

I hope this helps.

bakunin
# 11  
Old 09-16-2008
Bug solution
Quote:
Originally Posted by b_manu78
Hi Friends,

I see the last login time as

time_last_login=1210762918

How to convert this to standard format.

I believe there is a command, I am not able to recollect it.

Thanks in advance
do this.....

perl -le 'print scalar localtime('$a');'

and it returns

Wed May 21 07:48:33 2008
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