Need to filter out valid IP addresses using regex

Question

I have a Radius client configuration file in /etc/raddb/server in that want to get valid IP address without commented line,So I'm using

grep -o '[0-9]\{1,3\}\.[0-9]\{1,3\}\.[0-9]\{1,3\}\.[0-9]\{1,3\}' /etc/raddb/server
127.0.0.1
192.168.0.147

But I want to ignore 127.0.0.1 which is commented with # so how to do this stuff??

/etc/raddb/server file is as follow

cat /etc/raddb/server 
#  pam_radius_auth configuration file.  Copy to: /etc/raddb/server
#
#  For proper security, this file SHOULD have permissions 0600,
#  that is readable by root, and NO ONE else.  If anyone other than
#  root can read this file, then they can spoof responses from the server!
#
#  There are 3 fields per line in this file.  There may be multiple
#  lines.  Blank lines or lines beginning with '#' are treated as
#  comments, and are ignored.  The fields are:
#
#  server[:port] secret [timeout]
#
#  the port name or number is optional.  The default port name is
#  "radius", and is looked up from /etc/services The timeout field is
#  optional.  The default timeout is 3 seconds.
#
#  If multiple RADIUS server lines exist, they are tried in order.  The
#  first server to return success or failure causes the module to return
#  success or failure.  Only if a server fails to response is it skipped,
#  and the next server in turn is used.
#
#  The timeout field controls how many seconds the module waits before
#  deciding that the server has failed to respond.
#
# server[:port] shared_secret      timeout (s)
#127.0.0.1      secret             1
#other-server    other-secret       3
192.168.0.147:1812    testing123           1
#
# having localhost in your radius configuration is a Good Thing.
#
# See the INSTALL file for pam.conf hints.

Answer 1

try grep -o '^[0-9]\{1,3\}\.[0-9]\{1,3\}\.[0-9]\{1,3\}\.[0-9]\{1,3\}' /etc/raddb/server