bash - finding the tty of a logged in user -

i'm trying find tty number of person logging in using loops. code far is:

if [ "$#" -ne 1 ]   echo "usage: mon user"   exit 1 fi  user="$1"  #   until | grep "^$user " > /dev/null   tty=$(who | cut -c 9-13)   sleep 60 done  echo "$user has logged onto $tty" 

how go cutting out characters between 9-13 in first row? secondly, did correctly? can't test without having person work with.

you can reduce code lot loops not required seach extract user , tty

for example

if [ "$#" -ne 1 ] echo "usage: mon user" exit 1 fi  tty=$(who | grep "$user" | awk '{print $2}')  echo "$user has logged onto $tty" 

what does?

• who lists logged in users

• grep "$user" searches lines containing $user

• awk '{print $2}' awk prints second column in output

you can use cut instead of awk in program. instead of using -c option selects characters use -f option select fields

that like

 | grep "$user" | cut -d\  -f 2   

• -d\ sets delimtter space

• -f 2 selects second field output