select x.dbname as db_name, x.table_name as table_name, x.bytesize as the_size from
  (select
     table_schema as dbname,
     sum(index_length+data_length) as bytesize,
     table_name
   from
     information_schema.tables
   group by table_schema
  ) x
where
  x.bytesize > 999999
order by x.bytesize desc;

SELECT 
    table_name AS "Table",  
    round(((data_length + index_length) / 1024 / 1024), 2) as size   
FROM information_schema.TABLES  
WHERE table_schema = "YOUR_DATABASE_NAME"  
ORDER BY size DESC; 

这将对大小进行排序(DB大小以MB为单位)。

尝试以下shell命令(将DB_NAME替换为您的数据库名称):

mysql -uroot <<<"SELECT table_name AS 'Tables'， round((data_length + index_length) / 1024 / 1024)， 2)Size in MB FROM information_schematable WHERE table_schema = \"DB_NAME\" ORDER BY (data_length + index_length) DESC;"|头

对于Drupal/drush解决方案，检查下面的示例脚本，它将显示正在使用的最大表:

#!/bin/sh
DB_NAME=$(drush status --fields=db-name --field-labels=0 | tr -d '\r\n ')
drush sqlq "SELECT table_name AS 'Tables', round(((data_length + index_length) / 1024 / 1024), 2) 'Size in MB' FROM information_schema.TABLES WHERE table_schema = \"${DB_NAME}\" ORDER BY (data_length + index_length) DESC;" | head -n20

这只是一个供以后参考的说明。所有答案都依赖于I_S.TABLES。例如，如果你在表中有blob字段，它不会告诉你正确的大小。LOB页存储在外部页中，因此不计入聚集索引。 事实上有一个注释:

对于NDB表，该语句的输出显示适当的值 AVG_ROW_LENGTH和DATA_LENGTH列，例外 没有考虑到BLOB列。

我发现InnoDB也是如此。

我已经创建了社区Bug相同。

Size of all tables: Suppose your database or TABLE_SCHEMA name is "news_alert". Then this query will show the size of all tables in the database. SELECT TABLE_NAME AS `Table`, ROUND(((DATA_LENGTH + INDEX_LENGTH) / 1024 / 1024),2) AS `Size (MB)` FROM information_schema.TABLES WHERE TABLE_SCHEMA = "news_alert" ORDER BY (DATA_LENGTH + INDEX_LENGTH) DESC; Output: +---------+-----------+ | Table | Size (MB) | +---------+-----------+ | news | 0.08 | | keyword | 0.02 | +---------+-----------+ 2 rows in set (0.00 sec) For the specific table: Suppose your TABLE_NAME is "news". Then SQL query will be- SELECT TABLE_NAME AS `Table`, ROUND(((DATA_LENGTH + INDEX_LENGTH) / 1024 / 1024),2) AS `Size (MB)` FROM information_schema.TABLES WHERE TABLE_SCHEMA = "news_alert" AND TABLE_NAME = "news" ORDER BY (DATA_LENGTH + INDEX_LENGTH) DESC; Output: +-------+-----------+ | Table | Size (MB) | +-------+-----------+ | news | 0.08 | +-------+-----------+ 1 row in set (0.00 sec)

我发现现有的答案实际上并没有给出磁盘上表的大小，这更有帮助。 与基于data_length的表大小相比，此查询提供了更准确的磁盘估计 和索引。我不得不在AWS RDS实例中使用这种方法，因为您无法物理地检查磁盘和检查文件大小。

select NAME as TABLENAME,FILE_SIZE/(1024*1024*1024) as ACTUAL_FILE_SIZE_GB
, round(((data_length + index_length) / 1024 / 1024/1024), 2) as REPORTED_TABLE_SIZE_GB 
from INFORMATION_SCHEMA.INNODB_SYS_TABLESPACES s
join INFORMATION_SCHEMA.TABLES t 
on NAME = Concat(table_schema,'/',table_name)
order by FILE_SIZE desc