check if argument is an ip address in bash/sh

Shell Programming and Scripting
1

Hi all,

Can you please suggest a few lines of if statement to check if a variable is an ip address purely in bash/sh?

Thanks,
Marc

2

If you had searched the forum you would have found this post - IP address validation function

3

Well, I saw vino's reply only now that I've written my own function :slight_smile:
I post my code anyway:

IP="123.246.189.235"
TEST=`echo "${IP}." | /usr/xpg4/bin/grep -E "([0-9]{1,3}\.){4}"`

if [ "$TEST" ]
then
   echo "$IP" | nawk -F. '{
      if ( (($1>=0) && ($1<=255)) &&
           (($2>=0) && ($2<=255)) &&
           (($3>=0) && ($3<=255)) &&
           (($4>=0) && ($4<=255)) ) {
         print($0 " is a valid IP address.");
      } else {
         print($0 ": IP address out of range!");
      }
   }'
else
   echo "${IP} is not a valid IP address!"
fi

Is tested on Solaris, you may have to change the grep invocation according to your platform (grep, grep -E, egrep, ...).

4

thanks vino and robotronic