Ubuntu – Take user name as input: if user is logged in, display running processes belonging to the user

The main problem is here:

read $input

In bash, usually, $foo is the value of the variable foo. Here, you don't want the value, but the name of the variable, so it should be just:

read input

Similarly, in the if tests, $yes and $no should be just yes and no, since you just want the strings yes and no there.

You could use a case statement here, which (IMHO) makes it easier to do multiple cases based on input:

case $input in
[Yy]es)    # The first character can be Y or y, so both Yes and yes work
    echo "Hello!"  
    echo "Hello!" | festival --tts
    ;;
[Nn]o)     # no or No
    echo "Are you sure?"
    echo "Are you sure?" | festival --tts
    ;;
*)         # Anything else
    echo "Please answer yes or no."
    echo "Please answer yes or no." | festival --tts
    ;;
esac

You could wrap the two echo statements and the use of festival in a function to avoid repeating yourself:

textAndSpeech ()
{
    echo "$@"
    echo "$@" | festival --tts
}
case $input in
[Yy]es)    # The first character can be Y or y, so both Yes and yes work
    textAndSpeech "Hello!"  
    ;;
[Nn]o)     # no or No
    textAndSpeech "Are you sure?"
    ;;
*)         # Anything else
    textAndSpeech "Please answer yes or no."
    ;;
esac

With $input, bash replaces this with its value, which is nothing initially, so the read command run is:

read 

And read by default stores the input in the variable REPLY. So you can, if you want, eliminate the input variable altogether and use $REPLY instead of $input.

Also have a look at the select statement in Bash.

To modfy your current approach, you would need an array of counters for the different users, and an additional loop to test each value of $users against each positional parameter $1, $2, ... .

Note that for users in $(w -h) loops over all the whitespace-delimited tokens in the output of w -h, not just the usernames - it may "work" (because it's unlikely that other tokens match a valid username) but it's not "right" - you could use $(w -h | cut -d ' ' -f1) to extract just the usernames.

For a simpler approach, you could run w -h for each username given on the command line, and count the number of lines of output, for example using wc -l

#!/bin/sh

for user do

  counter=$(w -h "$user" | wc -l)

  if [ "$counter" -gt 2 ]; then
    printf "user %s is logged in %d times\n" "$user" "$counter"
  else
    printf "user %s is logged in less than two times\n" "$user"
  fi

done

Note that bash is not necessary for this - you can use the lighter /bin/sh shell (although the same code will run in bash).