shell_exec doesn't work with find

Question

I'm trying to implement php script on server created by someone else. I try echo shell_exec('ls'); and it works.

var_dump(shell_exec('find / -name "tmp.txt"'));

returns null. However when I try it in ssh console, the output is correct. What can cause it? What can I check

Answer 1

You're trying to locate something in the root directory, which your web server doesn't have permissions to access (by design) - what you're wanting to do would be a security concern, so my suggestion is to place your work in a folder inside your website, owned by the same user that is allowed to access files on your drive, typically www-data .

Perhaps you don't want to read the root directory of the server itself, and you just want to read a file in the same directory as the website. Switching out the / for a . will access the current directory, not the base directory:

var_dump(shell_exec('find . -name "tmp.txt"'));

Or, you could try an absolute path:

var_dump(shell_exec('find /path/to/files -name "tmp.txt"'));

This is uncommon, but you may need to point to an absolute path for find as well. You can test if you need to do this by running var_dump(shell_exec('find .')); and if it has no output then you will probably have to. Use /usr/bin/find instead if this is the case.

Read more about how to set permissions
Apache: File and Directory Ownership and Permissions for Web Content

Answer 2

A reason for this can be a missing PATH environment variable.

Please check getenv('PATH') and try to use the absolute path to find. A usual place is /usr/bin/find or /bin/find.

var_dump(shell_exec('/usr/bin/find / -name "tmp.txt"'));