How to get database names excluding system ones?

Question

MySQL

Wrote a MySQL dump script to files for myself. I know it's a bike, but for me to train. And here is the result on the screen:

Here is the script itself:

#!/bin/bash
#

DUMP_FOLDER=${PWD}/dump
DB_USER=root
DB_PASSWORD=root
DB_HOST=localhost

[ -d ${DUMP_FOLDER} ] || mkdir ${DUMP_FOLDER}

DB_LIST=$(echo "SHOW DATABASES;" | mysql --host=${DB_HOST} --user=${DB_USER} --password=${DB_PASSWORD})

for DB_NAME in $DB_LIST
do
    DB_FOLDER=${DUMP_FOLDER}/${DB_NAME}
    [ -d ${DB_FOLDER} ] || mkdir ${DB_FOLDER}
    TABLE_LIST=$(echo "USE ${DB_NAME};SHOW TABLES;" | mysql --host=${DB_HOST} --user=${DB_USER} --password=${DB_PASSWORD})
    for TABLE_NAME in $TABLE_LIST
    do
        mysqldump --host=${DB_HOST} --user=${DB_USER} --password=${DB_PASSWORD} -B ${DB_NAME} --tables ${TABLE_NAME} > ${DB_FOLDER}/${TABLE_NAME}.sql
    done
done

Actually, I only created the bashorg database (dump of quotes made using grab in python).
Service data, such as information_schema is not needed. The names of the required databases will be unknown in advance. How to exclude them?

Answer 1

DB_LIST=$(echo "SHOW DATABASES where \`database\` != 'mysql' AND \`database\` not like '%_schema';" | mysql --host=${DB_HOST} --user=${DB_USER} --password=${DB_PASSWORD})