Get only the subnet mask number on linux using bash

Question

I have tried this suggestion :

ip -o -f inet addr show | awk '/scope global/ {print $4}'

but that outputs IP address with the subnet mask number:

192.168.1.108/24

I only want the number 24

Answer 1

ip addr show can output JSON data, so it makes it reliably explicit to parse with jq:

ip \
  -family inet \
  -json \
  addr show |
    jq -r '.[].addr_info[0] | select(.scope == "global") | .prefixlen'

man ip:

-j, -json

Output results in JavaScript Object Notation (JSON).

Answer 2

With GNU awk, you may use gensub:

txt='scope global'
ip -o -f inet addr show | \
 awk -v search="$txt" '$0 ~ search{print gensub(/.*\//, "", 1, $4)}'

Here,

• -v search="$txt" - passes the value of txt to awk as search variable
• $0 ~ search - checks if there is a match in the whole line
• gensub(/.*\//, "", 1, $4) - removes all up to an including the last slash in the fourth field (replaces with an empty string (""), search is performed once only (1)).

Answer 3

this should only output the two or single digit subnet mask number like 24 :

ip -o -f inet addr show | grep -Po "/\K[[:digit:]]{1,2}(?=.*scope\sglobal)"

if you want it to output with the slash /24 :

ip -o -f inet addr show | grep -Po "/[[:digit:]]{1,2}(?=.*scope\sglobal)"

Answer 4

I used regex on your command to select everything after /

ip -o -f inet addr show | awk '/scope global/ {print $4}' | grep -o '[^/]*$'