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?
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
SELECT TABLE_NAME AS table_name,
table_rows AS QuantofRows,
ROUND((data_length + index_length) /1024, 2 ) AS total_size_kb
FROM information_schema.TABLES
WHERE information_schema.TABLES.table_schema = 'db'
ORDER BY (data_length + index_length) DESC;
以上2个都是在mysql上测试的
其他回答
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"
)
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)
有一种使用Workbench获取许多信息的简单方法:
右键单击模式名并单击“模式检查器”。 在生成的窗口中有许多选项卡。第一个标签 “信息”显示了数据库大小的粗略估计(以MB为单位)。 第二个选项卡“表”显示每个表的数据长度和其他详细信息。
如果您正在使用phpmyadmin,那么只需转到表结构
e.g.
Space usage
Data 1.5 MiB
Index 0 B
Total 1.5 Mi
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;