Obtaining list of files in all users directories

aarontwc · 12-17-2008, 03:15 AM

Let's say i have users: "aaron", "john", "peter" etc on a server, within which they have the file path "~/home/user/.prog/instances" individually.

As root, i am trying to run a cron job so that it reads all the files within "~/home/user/.prog/instances" and does something...

The question is, how do i obtain the list of files inside directory .prog/instances for all users?

I am certain this should be fairly simple, but I just can't find anything like this on Google

Aaron

jcookeman · 12-17-2008, 03:45 AM

Code:

for file in $(find /home/*/.prog/instances -type f -print)
do
    # something
    sed -i 's|foo|bar|g' $file
done

billymayday · 12-17-2008, 04:06 AM

Or maybe

find /home -type f -wholename '/home/*/.prog/instances/*' -exec dosomething \;

aarontwc · 12-18-2008, 09:23 PM

Hi there,

I am doing it like this:

Code:

my $cmd = "find /home/*/.prog/instances -type f -printf '%f\n' |";
open (MYFILES, $cmd) or die "Directory does not exist";
my @myFiles = <MYFILES>;
close MYFILES;

But, how do I create the directory if it does not exist?

I suppose the code is going to be:

Code:

if (!(-e $progDir)) {
    if (!mkdir($progDir,0755)) {
	die "$prog: Could not create $progDir.\n";
    }
}

How do i include that into the earlier code? So that I check for the directory, and if it does not exist, create it.

Thanks!!
Aaron

chrism01 · 12-21-2008, 08:08 PM

Looks like you want to sweep the dirs in Perl, so use opendir() http://perldoc.perl.org/functions/opendir.html.
Actually, Id start with the list of users and check '-e $wanted_dir', then opendir if exists, else mkdir...