SUID not working with shell script

Question

I have created a small shell script with the following content:

cat /usr/bin/checksuid.sh

!/bin/bash
echo \"Hello\" > /etc/myfile.cnf

ls -l /usr/bin/che         


        

Answer 1

From http://www.tuxation.com/setuid-on-shell-scripts.html:

"the truth is actually that the setuid bit is disabled on a lot of *nix implementations due the massive security holes it incurs"

An alternative approach - wrap the script in something that can use setuid, like this example c program. There are obviously differences to simply calling your script vs using a wrapper like this (e.g. ignored exit codes) but this should give you an idea anyway.

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>

int main()
{
   setuid( 0 );
   system( "/path/to/script.sh" );

   return 0;
}

Answer 2

Shell scripts can't be SUID. See http://www.faqs.org/faqs/unix-faq/faq/part4/section-7.html