Question

I am trying to set up a cron job that everyday optmizes the mysql database using debian on a virtual machine.

First I have tried using the mysqlcheck like this:

0 * * * * mysqlcheck -Aao –auto-repair -u root -p[password] > /dev/null

I got this error: mysqlcheck: Got error: 1045: Access denied for user ‘root’@’localhost’ (using password: YES) when trying to connect

after googling this error I found Dave Stoddard Comment who was experiencing the same problem:

0 * * * * /usr/local/bin/mysqlcheck --defaults-file=/root/.my.cnf --all-databases --auto-repair 2>&1 | grep -v "OK" | mail -s "Database Problem" root

or 0 * * * * /usr/local/bin/mysqlcheck --defaults-file=/root/.my.cnf --all-databases --auto-repair 2>&1 | grep -v "OK" | mail -s "Database Problem" root

Error from above:

/bin/sh: root: command not found
/bin/sh: /usr/local/bin/mysqlcheck: No such file or directory

Answer 1

seriously, you might over-doing,
your approach will causing some unnecessary table lock,
which mean it easily freezing your application that required database access

the optimization is required only when the table is too fragmented
meaning contains free-space,
an optimization able to impact the index key and data file

what is free-space?

this query will return list of table need to be optimized

select TABLE_SCHEMA, TABLE_NAME
from information_schema.tables 
where data_free>0;

build using a simple bash script

chk="select concat(TABLE_SCHEMA,'.', TABLE_NAME) from information_schema.tables where data_free>0;"

for tbl in $(mysql -u root -pxxx -N <<< $chk)
do
  mysql -u xxx -pxxx -N <<< "optimize table $tbl"
done