I need to get the IP from a log I need to grep the true-client and after that I need to grep true-client-ip=[191.168.171.15]
and get just the IP
2019.02.14-08:26:06:713,asd:1234:chan,0.000,asd,S,request-begin-site,POST,{remoteHost=1.2.3.4,remoteAddr=1.2.3.4,requestType=POST,serverName=api=[text/html],accept-charset=[iso-12345-15, utf-8;q=0.5, *;q=0.5],accept-encoding=[gzip],server-origin=[5],cache-control=[no-cache, max-age=0],pragma=[no-cache],program-header=[true],te=[chunked;q=1.0],true-client-ip=[191.168.171.15],true-host=[www.server.com]
I was trying grep -o "true-client-ip=[^ ]*,"
but it brings me:
true-client-ip=[191.168.171.15],true-host=[www.server.com]
I need just true-client-ip=[191.168.171.15]
so I can cut after to bring get the IP like true-client-ip=[191.168.171.15] | cut -d= -f2