Xrhrft

I am currently using information_schema.TABLES to calculate the total disk space usage grouped by the database name, but it is running terribly slowly. On servers with hundreds of databases, it can take minutes to calculate.

What is the quickest method of calculating disk space usage by database? Should I just be looking at the filesystem? Is there a method for speeding up information_schema?

There are 3 scenarios.

1. If you are using MyISAM, it is easiest to just look at the filesystem and use du -sh /var/lib/mysql/database.


2. If you are using InnoDB with innodb_file_per_table set, then you can get an approximate answer using du -sh. It is approximate because there is still some data stored in the ibdata1 file, so you will be a little on the low side. This technique also works with mixed MyISAM/InnoDB (innodb_file_per_table) databases.


3. If you are using InnoDB without innodb_file_per_table set, then you will need to look at INFORMATION_SCHEMA.

In any of the cases above, you can run the following query to get the information that you are looking for.

mysql> select table_schema, sum((data_length+index_length)/1024/1024) AS MB from information_schema.tables group by 1;

+--------------------+-----------------+

| table_schema       | MB              |

+--------------------+-----------------+

| prod               | 298025.72448921 |

| information_schema |      0.00781248 |

| maatkit            |     70.77330779 |

| mysql              |      0.66873168 |

| test               |   4752.31449127 |

+--------------------+-----------------+

5 rows in set (0.01 sec)

If you have a very large number of tables, it can be slow, as you have already discovered.

For getting info on the name of the table and number of records it has, the below query can be used,

SELECT * 

FROM information_schema.TABLES ;

For getting info on databases on the servers with their respective size, the below query can be used,

SELECT 

TABLE_SCHEMA  AS `Database`,

SUM((data_length + index_length) / (1024 * 1024)) AS `Database_Size`

FROM information_schema.TABLES 

GROUP BY table_schema 

ORDER BY `Database_Size` DESC;

In order for me to see where disk space is being used up (regardless if it's in a mysql table or not), I use my trusty "du" command. Here's an example of me finding where all the space is being eaten up from.

$ sudo du -cks /* | sort -rn

954881224   total

945218092   /mysql

5299904 /usr

1781376 /opt

1166488 /var

671628  /home

343332  /run

213400  /root

93476   /lib

30784   /boot

20652   /etc

15940   /bin

13708   /sbin

12388   /tmp

24  /mnt

16  /lost+found

4   /srv

4   /snap

4   /media

4   /lib64

0   /vmlinuz

0   /sys

0   /proc

0   /initrd.img

0   /dev

You can see that the majority of the space is being used by this folder.
/mysql

That folder holds data tables. In order to see which tables are taking all the space you can proceed like this using the "human" or "-h" option. I like to do disk space management this way because sometimes you cannot even log into mysql because you don't know the password or user.

$ sudo du -chs /mysql/*

2.3M    /mysql/blacklist

18M /mysql/clientservices

2.5G    /mysql/data

4.0K    /mysql/doubleverify

137G    /mysql/ias

4.0K    /mysql/IAS

2.2G    /mysql/innodb

16K /mysql/lost+found

4.0K    /mysql/ml_centroids

16G /mysql/moat

4.0K    /mysql/test

4.0K    /mysql/tmp

4.0K    /mysql/var

282G    /mysql/verticaAdFees

4.0K    /mysql/verticaViewability

247G    /mysql/Whiteops

217G    /mysql/Whiteops_TLX

902G    total

You can see that all space is being hogged by a few tables holding many GiG's of data. Hope this helps.

I would look for the size of the file on your data dictionnary. Its instantaneous and accurate.

Warning: According to the storage engine, indexes are stored within the main file or in another file don't forget to sum them up if needed.

I know this is old but someone may find this relevant.

In MySQL I use:

SELECT concat(table_schema) 'Database Name', concat(round(SUM(data_length/power(1024,3)),2),'G') DATA, concat(round(SUM(index_length/power(1024,3)),2),'G') 'INDEX', concat(round(SUM(data_free/power(1024,3)),2),'G') 'DATA FREE', concat(round(sum(data_free)/(SUM(data_length+index_length))*100,2)) '% FRAGMENTED', concat(round(SUM(data_length+index_length)/power(1024,3),2),'G') TOTAL FROM information_schema.TABLES WHERE table_schema NOT IN ('mysql','information_schema','performance_schema') GROUP BY table_schema;

Since my DB is InnoDB, this is just an estimate.

I compare this output to:

du -sch /location/of_Mysql/* | sort -hr | head -n20

Hope this helps you