create database using psql in shell script takes username as db name

Question

Following is the command I am using to create database in shell script

psql -v ON_ERROR_STOP=1 --username "$POSTGRES_USER" \
 -c "CREATE DATABASE $database;"

It throws error

FATAL: database "POSTGRES_USER" does not exist

If I print $database and $POSTGRES_USER they show correct values but when this command runs with some reason the username is passed as the database name.

any clue what I might be missing?

======

function create_db() {
 local database=1ドル
 local query_databases="select datname from pg_database;"
 local databases=$(echo "$query_databases" | psql -Aqt)
//this do not return any database but I have one db
 database_exist=$(containsElement "$databases" "$database")
 echo $database_exist;
 if [[ "$database_exist" == 1 ]]; then
 echo "Database '$database' exists. Skipping."
 else
 echo "Create Database '$database'"
 psql -v ON_ERROR_STOP=1 --username=$POSTGRES_USER \
 -c "CREATE DATABASE $database;"
 fi
}

taking reference from - https://github.com/mrts/docker-postgresql-multiple-databases/pull/10/files

• wildplasser: It is a default. If you don't supply a database name, postgres uses the username as a database name. So: you should supply a database name on the commandline. (postgres,or template1)
• user269867: @wildplasser shall I provide the name of the database which does not exists?
• wildplasser: You should supply a database name that does exist. (postgres and template1 (almost) always exist)
• user269867: That resolve the issue.

Answer 1

You need to remove the " in the command.

psql -v ON_ERROR_STOP=1 --username $POSTGRES_USER -c "CREATE DATABASE $database;"

• user269867: I still get the same error, I have updated with my function definition in the question.

Answer 2

You are not using psql terminal correctly. When using the long form of option, you need the = sign between the option and its value. Also the double quotes are not needed.

Consider:

psql -v ON_ERROR_STOP=1 --username=$POSTGRES_USER \
 -c "CREATE DATABASE $database;"

• user269867: I still get the same error, I have updated with my function definition in the question.