Regular expression using grep to extract an IP address

Question

I am trying to parse the default IP address of the default route.

I already have the default route and I'm trying to extract the IP address from it.

/sbin/ip addr show dev eth0 | grep 'inet'

Gets me as far as the correct line where the IP address is:

inet 10.1.4.33/22 brd 10.1.83.255 scope global eth0

And I need help extracting the IP address part 10.1.4.33

Answer 1

Pipe your output to grep -o :

/sbin/ip addr show dev eth0 | grep 'inet' | grep -oE "([0-9]{1,3}\\.){3}[0-9]{1,3}" | head -n 1

The head -n 1 is required to select the first match only.

Answer 2

You can use this awk :

/sbin/ip addr show dev eth0 | awk -F '[ /\t]+' '$2=="inet"{print $3; exit}'
192.168.0.52

Answer 3

尝试在awk中再尝试一种方法。

/sbin/ip addr show dev eth0 | awk '{match($0,/[0-9]+\.[0-9]+\.[0-9]+\.[0-9]+/);if(substr($0,RSTART,RLENGTH) && $0 ~ /inet/){print substr($0,RSTART,RLENGTH)}}'

Answer 4

要完成可用选项，请使用sed：

ip add show dev eth0 | sed -rn 's@^.*inet[[:blank:]]([[:digit:]]{1,3}(.[[:digit:]]{1,3}){3})/.*$@\1@p'

Answer 5

No complicated regular expression is needed.

output=$(/sbin/ip addr show dev eth0 | grep 'inet')
[[ $output = inet\ (.*)/ ]] && ip_addr=${BASH_REMATCH[1]}