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тАО06-08-2001 12:49 AM
тАО06-08-2001 12:49 AM
procedure to know the files accessed in a time window
How can i do a procedure that let me know the files accessed by a user in a time windiw ( 1 minute for example ) ?
Thanks
Federico
Thanks
Federico
3 REPLIES 3
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тАО06-08-2001 01:17 AM
тАО06-08-2001 01:17 AM
Re: procedure to know the files accessed in a time window
You can find find with more options here:
http://hpux.cs.utah.edu/hppd/hpux/Gnu/findutils-4.1.5/
I haven't tried it, but may provide something.
Other than that I can only suggest finding on two times individually and then diff'ing the output...
Later,
Bill
http://hpux.cs.utah.edu/hppd/hpux/Gnu/findutils-4.1.5/
I haven't tried it, but may provide something.
Other than that I can only suggest finding on two times individually and then diff'ing the output...
Later,
Bill
It works for me (tm)
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тАО06-08-2001 01:19 AM
тАО06-08-2001 01:19 AM
Re: procedure to know the files accessed in a time window
here's a gui based find
http://hpux.cs.utah.edu/hppd/hpux/Gtk/gtkfind-1.1/
and another general find tool
http://hpux.cs.utah.edu/hppd/hpux/Misc/findgrep-1.0/
Later,
Bill
http://hpux.cs.utah.edu/hppd/hpux/Gtk/gtkfind-1.1/
and another general find tool
http://hpux.cs.utah.edu/hppd/hpux/Misc/findgrep-1.0/
Later,
Bill
It works for me (tm)
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тАО06-08-2001 04:26 PM
тАО06-08-2001 04:26 PM
Re: procedure to know the files accessed in a time window
You could get an idea using lsof in a loop. Something like
(for i in 1 2 3 4 5 6 7 8 9 10 11 12; do
lsof | head +2
sleep 5
done) | cut -b 55- | sort -u
would give you a list of most files a user had open.
(for i in 1 2 3 4 5 6 7 8 9 10 11 12; do
lsof | head +2
sleep 5
done) | cut -b 55- | sort -u
would give you a list of most files a user had open.
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