Hi Jim,
Thanks for the reply. This seem to work only if i am running perl on the
machine itself.
I am instead using expect to ssh into this machine and run stat command and
use the output from that command to check the output of that stat command.
Thanks,
-Ben
On Tue, Mar 16, 2010 at 1:31 PM, Jim Gibson <jimsgibson@gmail.com> wrote:
> On 3/16/10 Tue Mar 16, 2010 12:18 PM, "ben perl" <ben.perl3@gmail.com>
> scribbled:
>
> > Hi Everyone,
> >
> > I writing this program to check if a file is being touched (linux "touch"
> > command) every 25 seconds.I am using stat command on linux.
> >
> > For example(please check the bold),
> >
> > stat file
> > File: `file'
> > Size: 0 Blocks: 0 IO Block: 4096 regular empty
> > file
> > Device: 803h/2051d Inode: 164394 Links: 1
> > Access: (0666/-rw-rw-rw-) Uid: ( 0/ admin) Gid: ( 0/ root)
> > Access: 2010-03-16 12:09:03.000000000 -0700
> > Modify: 2010-03-16 12:09:03.000000000 -0700
> > Change: *2010-03-16 *12:09:03.000000000 -0700 <=
>
>
> >
> > I am just worried about the change attribute above from stat command. So
> ,
> > what i am trying to do convert 12:09:03(as in the example above) into
> > seconds and add 25 seconds and convert back into "hour:minutes:seconds"
> > format and check the values match.
>
> If you are writing a Perl program, you can use the Perl built-in stat
> function that returns 'last modify time' in seconds from the epoch
> (1/1/70:00:00:00 UTC), rather than the Linux stat command as above. That
> way, you can easily compare times numerically and not worry about
> converting
> between numerical and string forms. See 'perldoc -f stat' for details.
>
> There is also the -M file test operator that returns the age (difference
> between the time your program started and the time the file was modified)
> of
> any file in floating-point days (multiply by 86,400 to get age in seconds).
> See 'perldoc -f -X' for details.
>
>
>
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>
>
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