<div class="gmail_quote">On Sat, Nov 21, 2009 at 17:23, Donald Russell <<a href="mailto:russell.don@gmail.com">russell.don@gmail.com</a>> wrote: <blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;"> I have a PHP web page that builds a list of files from a particular directory. In that PHP script I want to include some information about "what is using the file" (i.e ftp, gzip, cp etc.. so I code exec('/sbin/fuser -v $fpath/$fname',$answer,$retcode); Retcode is always 1 (not in use, or a "fatal error" happened) and $answer is always empty From root, in the directory I'm interested in I use fuser -v * and it shows me a couple of in-use items (for testing I have a little loop that gzip/gunzip a rather large file) I tried /sbin/fuser from my own id and get the same results as apache... that is, nothing useful. So I thought AH HA!, use sudo in the PHP script... I got an error message from sudo saying a tty is required to use sudo even though I had that command set up to not require password. Back to the drawing board... How can I issue a command from a PHP script that will "run as root" to get the fuser results for me? In some cases the file are chmod 600 root root in other cases 644 <regular user> <regular user> I want my web page thing to show me the files in-use... I can live without the 600 root-root files, but I really need to see the others. Any suggestions? Thank very much. </blockquote></div> Solution: $ chmod 6755 /sbin/fuser I'm not sure that's a really good idea.... but it gets the job done pretty nicely. I don't think that introduces any problems... only the program/command name is shown when using the -v option...