fetching a http process user using bash - regex

Question

Whats the best way to fetch a web process user (apache|nginx|www-data) for bash script usage?
In my case for setting up folder permissions and changing to the poper owner.
Currently I'm using:
ps aux | grep -E "(www-data|apache|nginx).*(httpd|apache2|nginx)" \
| grep -o "^[a-z\-]*" | head -n1
inside a bash script to fetch the owner of the http process.
Any hints on a more smartly solution or a better regex whould be great.

Answer 1

Your solution will really depend on your operating system. One option might be to check whether likely candidates exist in your password file:
user=$(awk -F: '/www|http/{print $1;exit}' /etc/passwd)
If you really want to look for the owner of running processes, remember that Apache often launches a root-owned "master" process, then launches children as the web user. So perhaps something like this:
user=$(ps aux|awk '$1=="root"{next} /www|http|apache/{print $1;exit}')
But you should also be able to determine things based on OS detection, since things tend to follow standards:
case "`uname -s`" in
Darwin) user=_www; uid=70 ;;
FreeBSD) user=www; uid=80 ;;
Linux)
if grep Ubuntu /etc/lsb-release; then
user=www-data; uid=$(id -u www-data)
elif [ -f /etc/debian_version ]; then
user=www-data; uid=$(id -u www-data)
elif etc
etc
fi
;;
esac
I'm not up on the best ways to detect different Linux distros, so that may require a bit of additional research for you.