How about using stat(2)?
$ ls -l proc.txt
-rw-rw-r-- 1 oracle dba 3414 Jul 1 00:10 proc.txt
$ perl -e '$a=(stat "proc.txt")[9]; print int $a/31536000+1970,"\n"'
2001
There's a better way to convert seconds since epoch to year but for now I just
divide it by number of seconds in a year and add 1970 to it.
The above perl one-liner is just a convenient way to call stat(2).
Yong Huang
[EMAIL PROTECTED]
you wrote:
But the year replaces the time in the 8th field only when the last
modification time for the file is more than 6 month (even if it is in the
current year :)
For example, take a look at line 1,2 (less than 6 month old as of today) &
3,4,5 (over 6 months old as of today)..
-rw-rw-r-- 1 oracle dba 2880 Feb 5 08:05 junk.lst
-rwxrwx--- 1 oracle dba 410 Jan 30 11:08 show_all.ksh
-rwx------ 1 oracle dba 77 Jan 25 2001 t1
-rw-rw-r-- 1 oracle dba 3971 Jan 10 2001 my.lst
-rw-rw-r-- 1 oracle dba 720 Jan 7 2001 bdf.out
HTH...
Regards,
- Kirti Deshpande
Verizon Information Services
http://www.superpages.com
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