我可以运行这个查询来获得MySQL数据库中所有表的大小:

show table status from myDatabaseName;

我希望有人能帮助我理解结果。我在找尺寸最大的桌子。

我应该看哪一列?


当前回答

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)

其他回答

你可以使用这个查询来显示表的大小(尽管你需要先替换变量):

SELECT 
    table_name AS `Table`, 
    round(((data_length + index_length) / 1024 / 1024), 2) `Size in MB` 
FROM information_schema.TABLES 
WHERE table_schema = "$DB_NAME"
    AND table_name = "$TABLE_NAME";

或者这个查询列出每个数据库中每个表的大小,最大的先:

SELECT 
     table_schema as `Database`, 
     table_name AS `Table`, 
     round(((data_length + index_length) / 1024 / 1024), 2) `Size in MB` 
FROM information_schema.TABLES 
ORDER BY (data_length + index_length) DESC;

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

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。

改编自ChapMic的回答,以满足我的特殊需要。

只指定数据库名称,然后按降序对所有表进行排序——在所选数据库中从最大到最小的表。只需要替换1个变量=数据库名。

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_HERE"
ORDER BY size DESC;

我使用这个shell脚本来跟踪表大小(以字节和行数为单位)

#!/bin/sh

export MYSQL_PWD=XXXXXXXX
TABLES="table1 table2 table3"

for TABLE in $TABLES;
do
        FILEPATH=/var/lib/mysql/DBNAME/$TABLE.ibd
        TABLESIZE=`wc -c $FILEPATH | awk '{print $1}'`
        #Size in Bytes
        mysql -D scarprd_self -e "INSERT INTO tables_sizes (table_name,table_size,measurement_type) VALUES ('$TABLE', '$TABLESIZE', 'BYTES');"
        #Size in rows
        ROWSCOUNT=$(mysql -D scarprd_self -e "SELECT COUNT(*) AS ROWSCOUNT FROM $TABLE;")
        ROWSCOUNT=${ROWSCOUNT//ROWSCOUNT/}
        mysql -D scarprd_self -e "INSERT INTO tables_sizes (table_name,table_size,measurement_type) VALUES ('$TABLE', '$ROWSCOUNT', 'ROWSCOUNT');"
        mysql -D scarprd_self -e "DELETE FROM tables_sizes WHERE measurement_datetime < TIMESTAMP(DATE_SUB(NOW(), INTERVAL 365 DAY));"
done

它假设有这个MySQL表

CREATE TABLE `tables_sizes` (
  `table_name` VARCHAR(128) NOT NULL,
  `table_size` VARCHAR(25) NOT NULL,
  `measurement_type` VARCHAR(10) NOT NULL CHECK (measurement_type IN ('BYTES','ROWSCOUNT')),
  `measurement_datetime` TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP()
) ENGINE=INNODB DEFAULT CHARSET=utf8