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
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”。
其他回答
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”。
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;
如果你有ssh访问权限，你可能想简单地尝试du -hc /var/lib/mysql(或不同的datadir，在my.cnf中设置)。
SELECT TABLE_NAME AS "Table Name",
table_rows AS "Quant of Rows", ROUND( (
data_length + index_length
) /1024, 2 ) AS "Total Size Kb"
FROM information_schema.TABLES
WHERE information_schema.TABLES.table_schema = 'YOUR SCHEMA NAME/DATABASE NAME HERE'
LIMIT 0 , 30
您可以从“information_schema”-> SCHEMATA表->“SCHEMA_NAME”列中获取模式名称
额外的 你可以得到mysql数据库的大小如下。
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
ORDER BY `DB Size in MB` DESC;
结果
DB Name | DB Size in MB
mydatabase_wrdp 39.1
information_schema 0.0
你可以在这里得到更多的细节。
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"
)
