Check if IP in array with Bash

I am trying to write a bash script which will take the output of who, parse it with awk pulling the IP address and then checking it against an array of IP addresses...sounds simple but they script I have put together does not seem to work, I have tested it on OSX and Ubuntu, can anyone spot why?

ip_address_is_safe() {
 local address_to_test=$1;
 for safe_ip in "${safe_ips[@]}"
 do
   if [[ $safe_ip == $address_to_test ]];
   then
     return 0;
   fi
 done;
 return 1;
}

who="root    pts/0        2011-10-03 23:13 (99.99.999.999)
root    pts/0        2011-10-03 23:13 (12.12.121.121)
root    pts/0        2011-10-03 23:13 (14.14.141.141)
root    pts/0        2011-10-03 23:13 (127.0.0.1)
";

safe_ips=("(14.14.141.141)" "(127.0.0.1)")


old_ifs=$IFS;
export IFS="
";
for word in $who; do
   remote_connected_ip=`echo $word | awk '/(23)/ {print $5}'`;
   if (ip_address_is_safe "$remote_connected_ip")
   then
       echo "ip was ok - $remot开发者_StackOverflowe_connected_ip"
   else
      echo "ip was not ok - $remote_connected_ip"
   fi
done;

It keeps reporting every IP as "ip was not ok"

Cheers!3

---

You're not quoting your variables, so awk is not seeing what you think it sees: you need

for word in "$who"; do
   remote_connected_ip=`echo "$word" | awk '/(23)/ {print $5}'`

This might be cleaner and does not require awk:

who | while read name tty date time ip; do
  if ip_address_is_safe $ip; then 
    echo $ip ok
  else
    echo $ip not ok
  fi
done  

---

if [[ $safe_id == ...

should be:

if [[ $safe_ip ==

Given awk is involved, you could do it all in awk:

who |
  awk 'BEGIN {
    n = split("(14.14.141.141) (127.0.0.1)", t)
    for (i = 0; ++i <= n;) safe_ips[t[i]]
    }
  $NF {
    print "ip was", ($NF in safe_ips ? x : "not"), "OK"
     }'