How do I drop all empty tables from a MySQL database, leaving only the tables that have at least 1 record?
EDIT: This is what I did using Python 3 with mysqlclient
import MySQLdb
conn = MySQLdb.connect(
host='localhost', user='root',
passwd='mypassword', db='mydatabase'
)
c = conn.cursor()
c.execute('SHOW TABLES')
for row in c.fetchall():
table_name = row[0]
c.execute(f'SELECT * FROM {table_name}')
if c.rowcount == 0:
# c.execute(f'DROP TABLE {table_name}')
print(f'DROP TABLE {table_name}')
Use pt-find from Percona Toolkit:
$ pt-find --dblike "mydatabase" --empty --exec-plus "DROP TABLE %s"
This stored procedure should do it:
DELIMITER $$
DROP PROCEDURE IF EXISTS `drop_empty_tables_from` $$
CREATE PROCEDURE `drop_empty_tables_from`(IN schema_target VARCHAR(128))
BEGIN
DECLARE table_list TEXT;
DECLARE total VARCHAR(11);
SELECT
GROUP_CONCAT(`TABLE_NAME`),
COUNT(`TABLE_NAME`)
INTO
table_list,
total
FROM `information_schema`.`TABLES`
WHERE
`TABLE_SCHEMA` = schema_target
AND `TABLE_ROWS` = 0;
IF table_list IS NOT NULL THEN
SET @drop_tables = CONCAT("DROP TABLE ", table_list);
PREPARE stmt FROM @drop_tables;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;
END IF;
SELECT total AS affected_tables;
END $$
DELIMITER ;
There may be problems with that GROUP_CONCAT
when there are too many empty tables. It depends on the value of the group_concat_max_len
system variable.
It's not possible to do it in one query because DROP TABLE
cannot receive its arguments from a SELECT
query.
Thanks to James for his comments. It appears that the row count query won't return precise results in the case of InnoDB tables, so the above procedure is not guaranteed to work perfectly when there are InnoDB tables in that schema.
For InnoDB tables, the row count is only a rough estimate used in SQL optimization. (This is also true if the InnoDB table is partitioned.)
Source: http://dev.mysql.com/doc/refman/5.1/en/tables-table.html
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