DATABASE autocreation bash script: What's wrong? - mysql

Question

My bash script won't crate the Database. What am I doing wrong here?
Please have a look:
#!/bin/bash -x
set -x
function deebee() {
EXPECTED_ARGS=2
E_BADARGS=65
MYSQL=`which mysql`
Q1="CREATE DATABASE IF NOT EXISTS $1;"
Q2="GRANT USAGE ON *.* TO $2#localhost IDENTIFIED BY '$3';"
Q3="GRANT ALL PRIVILEGES ON $1.* TO $2#localhost;"
Q4="FLUSH PRIVILEGES;"
SQL="${Q1}${Q2}${Q3}${Q4}"
if [ $# -ne $EXPECTED_ARGS ]
then
echo "Usage: $0 dbname dbuser dbpass"
exit $E_BADARGS
fi
$MYSQL -uroot -p -e "$SQL"
}
deebee $1 $2 $3
I'm calling the script as I've put it in a function, but it just spits out the expected arguments telling me the syntax, i.e that I should type in the bashscrip name, dbname, dbuser dbpass, but obviously there' something wrong with the script of my login permissions or user so that I can't automate this... What's going on, I'd love to know!
Thanks!

Answer 1

What I would do :
creating a mysql config file with credentials (easier to automate now) :
cat ~/.my.cnf
[client]
host = localhost
user = root
password = xxx
And the script :
#!/bin/bash
set -x
deebee() {
EXPECTED_ARGS=3
E_BADARGS=65
MYSQL=$(type -p mysql)
if (($# != $EXPECTED_ARGS))
then
echo "Usage: $0 dbname dbuser dbpass"
exit $E_BADARGS
fi
$MYSQL <<EOF
CREATE DATABASE IF NOT EXISTS '$1';
GRANT USAGE ON *.* TO '$2'#'localhost' IDENTIFIED BY '$3';
GRANT ALL PRIVILEGES ON '$1'.* TO '$2'#'localhost';
FLUSH PRIVILEGES;
EOF
}
# TODO tests on input args
deebee "$1" "$2" "$3"
And like Denis said in comments, take care of what users can put as arguments to prevent sql injection.
You should add some tests on input args.