Setup PHP Permission under Apache

Question

what i am trying to acomplish is to have a php page execute a shell command and display the output on the web page.

the code for the page is something along the lines of

<?php
echo shell_exec('ls -l /code');
?>

when i load up this page all i see is blank. yet when i execute it from the command line i see the command being executed and the output printed.

I am sure its something to do with the permissions and i am hoping someone can point me the right way.

this is what I see in the httpd log file

ls: cannot access /gateway_code: Permission denied
ls: cannot access /gateway_code: Permission denied

BTW the directory /code is set to "Read only" for the user "apache"

thanks! -ankit

Answer 1

Check your base directory restriction. Php doesn't have permission to access /bin/ls. Instead of adding /bin directory to the base directory restriction list copy /bin/ls over to the folder /code.

If you really have to do this.

Answer 2

BTW the directory /code is set to "Read only" for the user "apache"

That's your interpretation of the way the permissions have been set - it would have been a lot more useful if you'd told us what the permissions actually are.

I suspect the directory is not executable by the apache uid. Or there maybe MAC/ACL issues in play here - but you've provided no details of which Linux distribution this is, nor how the permissions are set up.

Answer 3

The first thing to do is make PHP give you a better answer than just a blank page. You can redirect stderr to stdout using 2>&1 such that the error message will be printed with echo.

<?php
echo shell_exec('ls -l /code 2>&1');
?>

Next, make sure that whatever process is running your php script (ie. apache) has execute permissions on the directory you're trying to list. If you don't mind being heavy-handed, you can easily determine whether permissions are the problem by just giving rwx access to everybody.

> chmod 777 code

And finally, make sure that you're specifying the right path. In your question you say that you're trying to list the contents of /code -- is it really your intention to be listing a root level directory? If the code folder is in the same directory as your script, you might have better luck with a relative path.

echo shell_exec('ls -l ./code 2>&1');