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)

这应该在mysql中测试，而不是postgresql:

SELECT table_schema, # "DB Name", 
Round(Sum(data_length + index_length) / 1024 / 1024, 1) # "DB Size in MB" 
FROM   information_schema.tables 
GROUP  BY table_schema; 

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

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)

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"
)

另一种显示所占用的行数和空间并按其排序的方法。

SELECT
     table_schema as `Database`,
     table_name AS `Table`,
     table_rows AS "Quant of Rows",
     round(((data_length + index_length) / 1024 / 1024/ 1024), 2) `Size in GB`
FROM information_schema.TABLES
WHERE table_schema = 'yourDatabaseName'
ORDER BY (data_length + index_length) DESC;  

在这个查询中唯一需要替换的字符串是“yourDatabaseName”。