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Shell script to find out owner, group permissions of file

Hi,
How do you write a shell script which takes an argument and determine what permissions the owner and group has and outputs the result in the following format

                                                                        READ            WRITE            EXECUTE

OWNER            User.Name                   YES                     YES                       NO
GROUP            GroupName                                YES                     NO                         NO

Thanks

F. Dominicus

Use stat on the files you are interested in or parse the output of ls -l on that file, if you have a printf in reach then use it. Or alternative use the templates from Perl.

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F. Dominicus

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wnross

You can use perl and stat to get the filemode information,
you can then mask out individual permissions:
so $mode & 0400 == user readable.

All of these permissions are listed as symbolic constants in the Fcntl package, the relevant ones are:
S_IRUSR S_IWUSR S_IXUSR
S_IRGRP S_IWGRP S_IXGRP
S_IROTH S_IWOTH S_IXOTH

Here's a code sample which would be trivial to finish: Save as perms.pl and try the following:
chmod +x perms.pl
./perms.pl /tmp
./perms.pl /var/adm
./perms.pl /home/myusername

--------------------- CUT HERE ---------------------
#!/usr/bin/perl
use Fcntl ':mode';

$file = <>;
chomp $file;

($mode,$uid,$gid) = (stat($file))[2,4,5];

$mode = $mode & 07777;
print "Mode $mode\nUID $uid\nGID $gid\n";
$usr = getpwuid($uid);
$grp = getgrgid($gid);

print "\n mode $mode " . S_IRUSR . "\n";
$u[4] = ($mode & S_IRUSR) >> 6;
$u[2] = ($mode & S_IWUSR) >> 6;
$u[1] = ($mode & S_IXUSR) >> 6;
$g[4] = ($mode & S_IRGRP) >> 3;
$o[1] = $mode & S_IXOTH;

print << "RESULTS"
READ WRITE EXECUTE
OWNER $usr $u[4] $u[2] $u[1]
GROUP $grp $g[4] $g[2] $g[1]
OTHER $o[4] $o[2] $o[1]

RESULTS

--------------------- CUT HERE ---------------------

Cheers,
-Bill