AIX

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:

Code:

perl -e 'print scalar localtime(1210762918);'

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..