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

if you have perl:

Thanks Yogesh,

How to get the cursor to the next line..

aixtech2:/>perl -e 'print scalar localtime(1210762918);'
Wed May 14 07:31:58 2008aixtech2:/>
aixtech2:/>

Try adding the -l option -- perl -le 'print ...'

Thanks a lot Yogesh.....Actually i am trying to write a shell script to find out the user accounts that were not used for more than for 4 months (based on last login) ..and lock them.

DO you have anything simpler than this ..

sorry ..it is Era...:-}

If i write script like below:

lastlog1=`lsuser -a time_last_login root|awk -F= '{print $2}'`
echo $lastlog1
perl -le 'print scalar localtime($lastlog1);'


Output is :

1211381313
Wed Dec 31 20:30:00 1969



It is not taking the correct date/time..

If i give perl -le 'print scalar localtime(1211381313);' , then the output is correct...Wed May 21 11:18:33 2008.

Please help me ...how can i pass the value as a variable and get the correct output ...i need to process all user accounts..

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

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.)

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

Just for the record: the problem is here:

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:

I hope this helps.

bakunin