Parse ifconfig to get only my IP address using Bash

Question

I want to edit the bashrc file to have a simple function called \"myip\" to run. As you might guess, the function myip prints only my internal IP address of my machine.

Answer 1

Taking patch's answer, making it a bit more general,

i.e.: skipping everything till the first digit.

ifconfig eth0 | awk '/inet addr/{sub(/[^0-9]*/,""); print $1}'

Or even better:

ifconfig eth0 | awk '/inet /{sub(/[^0-9]*/,""); print $1}'

• Please note the print part at the end - changes from $2 to $1.

Using Perl Regex:

ifconfig eth0 | grep -oP '(?<= inet addr:)[^ ]+'

Explanation: grep -oP searches for an EXACT match using Perl regex.
The "tricky" part is the regex itself;
1. (?<= inet addr:) means - that the string inet addr: is to the LEFT of what we're looking for.
2. [^ ]+ (please notice the space after the ^ ) - it means to look for everything until the first blank - in our case it is the IP Address.