ShellScript的PSQL COPY

Question

I am writing a shell script that fetches data (.csv file) form AWS S3, downloads it locally onto an EC2 Linux AMI Instance, and then copies the data to an RDS PostGresql database.

My Shell code is the following:

FILE="$(ls DB)"
PARAMETERFORDB= "'\\COPY table(x,y) FROM ''$FILE'' CSV HEADER'"

$(psql --host=XXXXX --port=XXXXX --username=XXXXX --password --dbname=XXXXX -c ${PARAMETERFORDB})

So when the data from S3 is downloaded, I store the files' name inside the FILE variable (it is the only file in the folder, the folder will be deleted after the Database query).

I get following error message:

./shellTest.sh: line 21: '\COPY table(x,y) FROM ''14.9.2016.csv'' CSV HEADER': command not found
psql: option requires an argument -- 'c'
Try "psql --help" for more information.

What am I doing wrong?

Answer 1

In the line

PARAMETERFORDB= "'\\COPY table(x,y) FROM ''$FILE'' CSV HEADER'"

remove the space after the = and remove one level of single quotes:

PARAMETERFORDB="\\COPY table(x,y) FROM '$FILE' CSV HEADER"

In the line where psql is invoked, enclose ${PARAMETERFORDB} in double quotes since it contains spaces.