Escaping variables in bash file - linux

Question

I'm creating a init.d script within a bash file, which goes as follows:
# AUTOSTART
$APPDIR=somedir
$APPCONF=somedir
$APPVENV=somedir
$APPUSER=someuser
cat <<EOF >/etc/init.d/uwsgi
#!/bin/bash
daemon=$APPVENV/bin/uwsgi
args="--emperor $APPCONF/uwsgi/app.ini --daemonize /var/log/emperor.log --emperor-pidfile $APPDIR/emperor.pid --gid `id -g $APPUSER`"
pid=$APPDIR/emperor.pid
case "$1" in
start)
echo "Starting uwsgi"
start-stop-daemon -m -p $pid --start --exec $daemon $args
;;
stop)
echo "Stopping script uwsgi"
start-stop-daemon --signal INT -p $pid --stop $daemon $args
;;
reload)
echo "Reloading conf"
kill -HUP $(cat $pid)
;;
*)
echo "Usage: /etc/init.d/uwsgi {start|stop|reload}"
exit 1
;;
esac
exit 0
EOF
It is my understanding, from help, that $APPCONF, $APPVENV, $APPUSER and $APPDIR need to be escaped because I define them outside the file. So is it correct that I simply put a back slash in front of the variable like this:
daemon=\$APPVENV/bin/uwsgi
args="--emperor \$APPCONF/uwsgi/app.ini --daemonize /var/log/emperor.log --emperor-pidfile \$APPDIR/emperor.pid --gid `id -g \$APPUSER`"
pid=\$APPDIR/emperor.pid
It still doesn't seem to work though, the service doesn't start, so I think I might have done something else wrong. Can anyone confirm I am escaping properly please?

Answer 1

The dollar sign is only used for reading/using the value of a variable, not when setting its value. Thus, setting APPDIR to somedir would look like this:
APPDIR=somedir
Depending on how you call the script, you may also want to export the variable:
export APPDIR=somedir