Extract IP address from a file - grep

Question

I have a log file where between the lines we have strings like " ,true-client-ip=[1.1.1.1]". I want to write a grep/sed command to extract the IPs from the file. What is the best way to extract this output and write it to another file?
cache-control=[no-cache, max-age=0],origin=[https://www.example.com],pragma=[no-cache],te=[chunked;q=1.0],true-client-ip=[108.81.148.189],user-agent=[Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.36 Edge/16.16299]

Answer 1

You can use expression:
([0-9]{1,3}\.){3}[0-9]{1,3}
You can write the output of the grep command to a new file as follows:
grep -Eo '([0-9]{1,3}\.){3}[0-9]{1,3}' yourlogfile.txt >> ipfile.txt
A second option would be to use the Perl dialect, and to use a lookbehind to extract the IP after the true-client-ip field.
grep -Po '(?<=true-client-ip=\[)[^\]]+' yourlogfile.txt >> ipfile.txt
ipfile.txt now contains:
108.81.148.189