Extracting IP address from a line from ifconfig output with grep

Question

Given this specific line pulled from ifconfig, in my case:

inet 192.168.2.13 netmask 0xffffff00 broadcast 192.168.2.255

How could one extract the 192.168.2.13 part (the local IP address), presumably with regex?

Answer 1

Here's one way using grep:

line='inet 192.168.2.13 netmask 0xffffff00 broadcast 192.168.2.256'  echo "$line" | grep -oE "\b([0-9]{1,3}\.){3}[0-9]{1,3}\b" 

Results:

192.168.2.13 192.168.2.256 

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If you wish to select only valid addresses, you can use:

line='inet 192.168.0.255 netmask 0xffffff00 broadcast 192.168.2.256'  echo "$line" | grep -oE "\b((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\b" 

Results:

192.168.0.255 

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Otherwise, just select the fields you want using awk, for example:

line='inet 192.168.0.255 netmask 0xffffff00 broadcast 192.168.2.256'  echo "$line" | awk -v OFS="\n" '{ print $2, $NF }' 

Results:

192.168.0.255 192.168.2.256 

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Addendum:

Word boundaries: \b

Answer 2

use this regex ((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)(?=\s*netmask)