最后计算数据库的总大小:

(SELECT 
  table_name AS `Table`, 
  round(((data_length + index_length) / 1024 / 1024), 2) `Size in MB` 
  FROM information_schema.TABLES 
  WHERE table_schema = "$DB_NAME"
)
UNION ALL
(SELECT 
  'TOTAL:',
  SUM(round(((data_length + index_length) / 1024 / 1024), 2) )
  FROM information_schema.TABLES 
  WHERE table_schema = "$DB_NAME"
)

如果您想查询使用当前选定的数据库。简单地复制粘贴这个查询。(无需修改)

SELECT table_name ,
  round(((data_length + index_length) / 1024 / 1024), 2) as SIZE_MB
FROM information_schema.TABLES
WHERE table_schema = DATABASE() ORDER BY SIZE_MB 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

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)

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。