Trying to capture IP address from `ifconfig` in terminal

Question

I'm trying to come up with a quick command in order to get the IP address of a computer via grep and the ifconfig command.

So far, I have this ifconfig eth0 | grep -Eo 'inet addr:[0-9\\.]+' ifconfig eth0 | grep -Eo 'inet addr:[0-9\\.]+' which will return:

inet addr:192.168.1.26

I'm trying to adjust the regex though so that I only get the IP address itself. I have very limited knowledge of regex, and I put a non-capturing group around 'inet addr:' which made the command look like this:

ifconfig eth0 | grep -Eo '(?:inet addr:)[0-9\\.]+'

but that still didn't solve my issue.

Answer 1

You can use:

ifconfig | awk '$1=="inet" && $2!="127.0.0.1"{print $2}'

Or on Linux:

ifconfig eth0 | awk -F '[ :]+' '$2=="inet" {print $4}'

Answer 2

You can use cut , to cut output into substrings by delimiter : .

    ifconfig eth0 | grep -Eo 'inet addr:[0-9\.]+' | cut -d':' -f 2

Flag -f 2 is used to select second substring.

Answer 3

Use a lookbehind expression; with grep , you need to add the -P switch to enable this construct instead of -E (at least under bash v4.1.1 for cygwin). And so just change a little your last try:

ifconfig eth0 | grep -Po '(?<=inet addr:)[0-9\.]+'