I have a working script for login time, but if the user isn't logged on I get an error

I have a working script that tells how long as user has been logged on . But it the user isn't logged on it will say they are logged off but it will give me an error for one of the tests in the middle of the script.

Im trying to figure out how to skip these lines if the user isn't online
if [ $eMin -lt 0 ] then eMin=$(expr $eMin + 60 ) eHr=$(expr $eHr - 1 ) fi

This is the full script for context

 while [ -z "$1" ]
 do
echo -n "Please enter valid id: "
read var1
set $var1 $1
# Break the loop
if [ -n "$1" ]
then
break
fi
done

# Check if the user is a valid user
if id "$1" >/dev/null 2>1;
# If so, then check if he if he is currently logged on and set on/not on 
as 
variable
then
# Set full name variable based on matching ID
fullname=$(grep "$1" /etc/passwd | cut -d ':' -f5 | sort -k 2 | tr ",,:" " 
" | 
awk '{print $2,$1}')
# Get Current  Hours and Minutes
nowHr=$(date | cut -c 12,13)
nowMin=$(date | cut -c 15,16)
onHr=$(who |grep "$1" | cut -c 34,35)
onMin=$(who |grep "$1" |cut -c 37,38)
# Hours minutes spent logged on
eHr=$(expr $nowHr - $onHr )
eMin=$(expr $nowMin - $onMin )

if [ $eMin -lt 0 ]
then
eMin=$(expr $eMin + 60 )
eHr=$(expr $eHr - 1 )
fi

# Test and display user name and if curretnly logged on
# Exit with code of 0 if success and 1 if fail
who -u |grep -q "$1" || test && echo "$fullname is logged on for $eHr 
hour(s) 
and $eMin minutes(s)." &&  exit 0 || echo "$fullname is not logged on" && 
exit 
1
# Displays if invalid id is entered
else
echo "The user you entered, $1 is not a valid user on this system"
# Exit with code of 2
fi
exit 2

The error message is expr: syntax error expr: syntax error ./timeOn: line 38: [: -lt: unary operator expected John Doe is not logged on