Fetching wrong results on comparing a variable with 0 in unix

I have the following code in unix #!/bin/bash user=$1 username=$user'%' DB_name=$2 userpwd=$4 export PGPASSWORD=$3 psql -h localhost -U postgres -d $DB_name -At -c "SELECT COUNT(*) AS USER_COUNT FROM pg_roles WHERE rolname = '${user}' " \ | while read -a Record ; do USER_COUNT=${Record} echo 'user USER_COUNT ' $USER_COUNT done echo 'count' $count if [[ $(($USER_COUNT)) -eq 0 ]] then echo 'New User ' $user else echo 'User Exists ' $user fi It basically accepts a given username and checks if it exists in Postgresql. After we pass the arguments, it checks its count, which will be 1, if the username exists, otherwise 0, and stores it in the variable USER_COUNT. It then compares it with 0. But the comparison is fetching me wrong results. Regardless of whether a username exists or not, it only runs the 'True' part of if. _Edit-1_ The reason I used if [[ $(($USER_COUNT)) -eq 0 ]] instead of if [$USER_COUNT -eq 0 ], as suggested by Phill, is because I was getting an error message otherwise, saying unary operator expected. I read somewhere that it is because $USER_COUNT is an unset variable. _Edit-2_ Thanks for your suggestion Lennart. As suggested by him, I added -x to my first line and found that even though the value of $USER_COUNT is 1 when the user exists, and 0 otherwise, but while comparison, i.e in the step if [[ $(($USER_COUNT)) -eq 0 ]], $USER_COUNT takes the value 0, even if earlier it was 1. Hence, it always evaluates to true. But I am not sure what to do to resolve the above situation. _Edit-3_ Thanks a lot Lennart. Your code solved my problem.