Which terminal command to get just IP address and nothing else?

Question

I\'m trying to use just the IP address (inet) as a parameter in a script I wrote.

Is there an easy way in a unix terminal to get just the IP address, rather than loo

Answer 1

#!/bin/sh
# Tested on Ubuntu 18.04 and Alpine Linux 
# List IPS of following network interfaces:
# virtual host interfaces
# PCI interfaces
# USB interfaces
# ACPI interfaces
# ETH interfaces
for NETWORK_INTERFACE in $(ls /sys/class/net -al | grep -iE "(/eth[0-9]+$|vif|pci|acpi|usb)" | sed -E "s@.* ([^ ]*) ->.*@\1@"); do 
    IPV4_ADDRESSES=$(ifconfig $NETWORK_INTERFACE | grep -iE '(inet addr[: ]+|inet[: ]+)' | sed -E "s@\s*(inet addr[: ]+|inet[: ]+)([^ ]*) .*@\2@")
    IPV6_ADDRESSES=$(ifconfig $NETWORK_INTERFACE | grep -iE '(inet6 addr[: ]+|inet6[: ]+)' | sed -E "s@\s*(inet6 addr[: ]+|inet6[: ]+)([^ ]*) .*@\2@")
    if [ -n "$IPV4_ADDRESSES" ] || [ -n "$IPV6_ADDRESSES" ]; then
        echo "NETWORK INTERFACE=$NETWORK_INTERFACE"
        for IPV4_ADDRESS in $IPV4_ADDRESSES; do 
            echo "IPV4=$IPV4_ADDRESS"
        done
        for IPV6_ADDRESS in $IPV6_ADDRESSES; do 
            echo "IPV6=$IPV6_ADDRESS"
        done
    fi
done