sh/expect/sudo/screen all needed together in one command.

Question

What I need to have happen: PHP needs to launch an server app which has root permissions running in the background. All of this should be silent.

-Sudo is needed to allow php to perform an op that requires root permissions.

-Screen is required to allow the app to run outside the scope of the webpage which started the process.

-Expect is needed so that screen has a pts in which to run

-Sh is needed because whatever starts running needs to be backgrounded, presumably with the & operator. It would also need to pipe any output to /dev/null/since I don't want my PHP page returning anything. This is probably negotiable somehow if somebody can think of a better way to do the call in PHP (fork...?)

As an example, the script I tried to use was:

#!/usr/bin/expect -f
spawn sh ( screen -t srcds /usr/local/srcds_l/startserv )& > /dev/null
exit 0

For reference I am trying to start a Counter-Strike Source server, and startserv is the name of the C code which handles launching the server and collecting its output. Can anybody correct my syntax for that snippet, or tell me why its the wrong thing to do?

Answer 1

I don't have a complete solution for you, but you might want to look at nohup instead of screen. That eliminates the Expect need as well.

Answer 2

That's a huge amount of overhead to communicate 1 bit of information ("start yourself now"). If you forget PHP then all the other subordinate requirements fall away.

If I was trying to do this, I'd probably write a perl script to listen on a TCP port for a "start" command and just fork off the process itself. This gets rid need for sh, expect, screen, and sudo.