SUID/GUID files search !

[fly over]

Thanks Paul for supporting, yes it is a sort of assignment but from my boss.
as your script returning file name, Please tell me how can i get the 
absolute pathname of the file and the owner.

Message: 17
Date: Tue, 01 Feb 2005 11:18:34 +0000
From: Paul Howarth <paul at city-fan.org>
Subject: Re: SUID/GUID files search !
To: For users of Fedora Core releases <fedora-list at redhat.com>
Message-ID: <41FF658A.8000506 at city-fan.org>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed

fly over wrote:
> Hi guys, i'm trying to write a script for following purpose.
>
> shell program will be used by Linux/Unix sysadmins to search for 
> SUID/SGID files. The default directory to search is the present 
> working directory, however, the user may include a directory name 
> on the command line as an alternative. Also, if the user includes 
> the argument '-R' then the search should include all subdirectories 
> recursively. Also, the '-G' argument will include SGID files which 
> by default are not shown. The output of the script should show the 
> absolute pathname of the file and the owner.
>
>
> Please help me in performin such task.

Sounds like a homework assignment to me...

> i'm trying using this line:
>
> ls -l | awk '{print $1}' | grep s
> it just prints the permissions having s bit.

Yes, that's right. Your awk command is just printing the first field of
the ls -l output, which is the permissions. What you want to do is to
search the permissions but output the filename, something more like this:

$ ls -l | awk '/^-..[Ss]/ { print $9 }'

This looks for regular files (1st character of line is "-") that have
the SUID bit set (fourth character of line is "s" or "S") and then
prints out the filename (9th field of line).

You probably want to be using the "find" command rather than the "ls"
command though. Use "-maxdepth 1" by default to turn off its recursive
checking of directories, and skip the "-maxdepth 1" option when your
script is passed the -R option.

Paul.
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