Linux bash script to extract IP address

Question

I want to make big script on my Debian 7.3 ( something like translated and much more new user friendly enviroment ). I have a problem. I want to use only some of the informa

Answer 1

If the goal is to find the IP address connected in direction of internet, then this should be a good solution.

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UPDATE!!! With new version of linux you get more information on the line:

ip route get 8.8.8.8
8.8.8.8 via 10.36.15.1 dev ens160 src 10.36.15.150 uid 1002
    cache

so to get IP you need to find the IP after src

ip route get 8.8.8.8 | awk -F"src " 'NR==1{split($2,a," ");print a[1]}'
10.36.15.150

and if you like the interface name

ip route get 8.8.8.8 | awk -F"dev " 'NR==1{split($2,a," ");print a[1]}'
ens192

ip route does not open any connection out, it just shows the route needed to get to 8.8.8.8. 8.8.8.8 is Google's DNS.

If you like to store this into a variable, do:

my_ip=$(ip route get 8.8.8.8 | awk -F"src " 'NR==1{split($2,a," ");print a[1]}')

my_interface=$(ip route get 8.8.8.8 | awk -F"dev " 'NR==1{split($2,a," ");print a[1]}')

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Why other solution may fail:

ifconfig eth0

• If the interface you have has another name (eno1, wifi, venet0 etc)
• If you have more than one interface
• IP connecting direction is not the first in a list of more than one IF

Hostname -I

• May get only the 127.0.1.1
• Does not work on all systems.