Passing Local IP as argument when running command line application in Unix - unix

Question

I have a command line application which I use and also have to pass my local ip address as an argument, like:
jekyll --url 'http://192.168.1.2:3000' --pygments --safe --server 3000 --auto
I would like to make the url argument get my ip automatically, since I am always on different networks and get different loal ip addresses.
so I can use this alias in my .bashrc
alias jkl="jekyll --url 'http://$IP:3000' --pygments --safe --server 3000 --auto"
where $IP would be my local ip adress acquired dynamically.
Is there any way to do it?

Answer 1

First, use double quotes instead of single quotes around your $IP variable or else it won't interpolate the value
#!/bin/bash
# tested on bash 4
while read -r line
do
case "$line" in
"inet "* )
line="${line/inet /}"
line="${line%% *}"
if [[ ! $line =~ ^(127|172) ]] ;then
IP="$line"
echo "IP: $IP"
fi
;;
esac
done < <(ifconfig)
echo jekyll --url "http://$IP:3000" --pygments --safe --server 3000 --auto
Note that you will have a few different IPs in the output. Choose the one that fits your requirement most.

Answer 2

A computer does not necessarily have "a local IP address", there are often several. For instance, you typically have the localhost address (127.0.0.1), and one or more "true" externally visible addresses. It's hard for an automated solution to know which one to pick.
One easy solution is perhaps to hard-code the "eth0" interface (or whatever the name is of your most typical interface).
On Linux, you could use something like this:
$ ifconfig | grep -A1 eth0 | cut -d: -f2 | cut -d ' ' -f1 | grep \\.
192.168.0.8
So to stuff this into a variable (assuming bash) you would use
MY_IP=$(ifconfig | grep -A1 eth0 | cut -d: -f2 | cut -d ' ' -f1 | grep \\.)
Note that this hard-codes the interface name as eth0.