regex - 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 "([0-9]{1,3}.){3}[0-9]{1,3}"

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 "((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]?)"

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="
" '{ print $2, $NF }'

Results:

192.168.0.255
192.168.2.256

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Addendum:

Word boundaries: