How to run MySQL command on bash?

Use double quotes while using BASH variables.

mysql --user="$user" --password="$password" --database="$database" --execute="DROP DATABASE $user; CREATE DATABASE $database;"

BASH doesn't expand variables in single quotes.

This one worked, double quotes when $user and $password are outside single quotes. Single quotes when inside a single quote statement.

mysql --user="$user" --password="$password" --database="$user" --execute='DROP DATABASE '$user'; CREATE DATABASE '$user';'

I have written a shell script which will read data from properties file and then run mysql script on shell script. sharing this may help to others.

#!/bin/bash    PROPERTY_FILE=filename.properties    function getProperty {       PROP_KEY=$1       PROP_VALUE=`cat $PROPERTY_FILE | grep "$PROP_KEY" | cut -d'=' -f2`       echo $PROP_VALUE    }    echo "# Reading property from $PROPERTY_FILE"    DB_USER=$(getProperty "db.username")    DB_PASS=$(getProperty "db.password")    ROOT_LOC=$(getProperty "root.location")    echo $DB_USER    echo $DB_PASS    echo $ROOT_LOC    echo "Writing on DB ... "    mysql -u$DB_USER -p$DB_PASS dbname<<EOFMYSQL    update tablename set tablename.value_ = "$ROOT_LOC" where tablename.name_="Root directory location";    EOFMYSQL    echo "Writing root location($ROOT_LOC) is done ... "    counter=`mysql -u${DB_USER} -p${DB_PASS} dbname -e "select count(*) from tablename where tablename.name_='Root directory location' and tablename.value_ = '$ROOT_LOC';" | grep -v "count"`;    if [ "$counter" = "1" ]    then    echo "ROOT location updated"    fi