UNIX for Dummies Questions & Answers

The first line is:

#!/usr/bin/sh

and just to see that it's there:

$ls -la /usr/bin/sh
-r-xr-xr-x 2 bin bin 208896 Oct 3 2004 /usr/bin/sh


Any ideas what I'm doing wrong??

Thanks for the help!

In order for your shell to execute a script, it has to be able to read it during the execution - it is parsed, not compiled.

There are a few work arounds... You can use the setuid bit, so that anyone that executes it, does it as you (or a general account) that owns the file. For example, if I have a shell script that echos "It works":

Code:

# chmod 4501 ick
# ls -l ick
-r-s-----x   1 myuser mygroup 22 Jan 18 17:03 ick
# su - notmyuser
$ ./ick
It works!

The users can now execute the sciprt, but will not be able to view it. You could also limit it to a group instead of having the other bit be executable. The bad thing is that the script then executes as you - a reason a general account might be better for ownership of the file.

If you are using Solaris, you can use profile shell and the /etc/security/* files to setup a better option - I'll cover that if you are actually using Solaris.

Cheers,

Keith

If the process cannot read the file, how can the process execute?

Actually I'm building this example on HP-UX.

You mention that the files are parsed, not compiled. Is it possible to compile a shell script so that they can only view and execute the compiled code, not the actual source code?