Beginner's Bash Scripting: “Unary Operator Expected” error && how to use grep to search for a variable from the output of a command?

Question

I'm attempting to write a bash script that is executed through "./filename username" and checks whether or not that user is logged in, printing the result. I'm still new to scripting and am having trouble understanding how to make this work.

I'm currently getting the error "line 7: [: ambonill: unary operator expected". What does that mean and how can I go about fixing that error?

Additionally, how would I get grep to work instead of sort | uniq? I'd like to grep for the variable from the output of the command but can't find anything related in the man page.

#! /bin/bash
# This script will take a username as an argument and determine whether they are logged on.

function loggedin {
   for u in `who | cut -f1 -d" " | sort | uniq`
   do
   if [ $u == $1 ]
   then
      echo "$1 is logged on"
   else
      echo "$1 is not logged on"
   fi
   exit 0
   done
}

loggedin $u
exit 1

Answer 1

Try to find a simpler solution, like:

#!/bin/bash
echo "$1 is $([ -z "$(w -h $1)" ]&&echo -n not\ )logged on"