我继承了一个相当大的SQL Server数据库。考虑到它包含的数据,它似乎比我预期的要占用更多的空间。
是否有一种简单的方法来确定每个表占用的磁盘空间?
我继承了一个相当大的SQL Server数据库。考虑到它包含的数据,它似乎比我预期的要占用更多的空间。
是否有一种简单的方法来确定每个表占用的磁盘空间?
当前回答
在上面的@Mark回答中,添加了@updateusage='true'以强制更新最新的尺码统计(https://msdn.microsoft.com/en-us/library/ms188776.aspx):
SET NOCOUNT ON
DECLARE @TableInfo TABLE (tablename varchar(255), rowcounts int, reserved varchar(255), DATA varchar(255), index_size varchar(255), unused varchar(255))
DECLARE @cmd1 varchar(500)
SET @cmd1 = 'exec sp_spaceused @objname =''?'', @updateusage =''true'' '
INSERT INTO @TableInfo (tablename,rowcounts,reserved,DATA,index_size,unused)
EXEC sp_msforeachtable @command1=@cmd1
SELECT * FROM @TableInfo ORDER BY Convert(int,Replace(DATA,' KB','')) DESC
其他回答
从使用OSQL的命令提示符:
OSQL -E -d <*databasename*> -Q "exec sp_msforeachtable 'sp_spaceused [?]'" > result.txt
以上查询有助于查找表(包括索引)使用的空间量,但如果要比较表上的索引使用了多少空间,请使用以下查询:
SELECT
OBJECT_NAME(i.OBJECT_ID) AS TableName,
i.name AS IndexName,
i.index_id AS IndexID,
8 * SUM(a.used_pages) AS 'Indexsize(KB)'
FROM
sys.indexes AS i
JOIN sys.partitions AS p ON p.OBJECT_ID = i.OBJECT_ID AND p.index_id = i.index_id
JOIN sys.allocation_units AS a ON a.container_id = p.partition_id
WHERE
i.is_primary_key = 0 -- fix for size discrepancy
GROUP BY
i.OBJECT_ID,
i.index_id,
i.name
ORDER BY
OBJECT_NAME(i.OBJECT_ID),
i.index_id
当处理多个分区和/或筛选索引时,Marc_s的答案给出了错误的结果。它也不区分数据和索引的大小,这通常是非常相关的。一些建议的修复方法并不能解决核心问题,或者根本就是错误的。
以下查询解决了所有这些问题。
SELECT
[object_id] = t.[object_id]
,[schema_name] = s.[name]
,[table_name] = t.[name]
,[index_name] = CASE WHEN i.[type] in (0,1,5) THEN null ELSE i.[name] END -- 0=Heap; 1=Clustered; 5=Clustered Columnstore
,[object_type] = CASE WHEN i.[type] in (0,1,5) THEN 'TABLE' ELSE 'INDEX' END
,[index_type] = i.[type_desc]
,[partition_count] = p.partition_count
,[row_count] = p.[rows]
,[data_compression] = CASE WHEN p.data_compression_cnt > 1 THEN 'Mixed'
ELSE ( SELECT DISTINCT p.data_compression_desc
FROM sys.partitions p
WHERE i.[object_id] = p.[object_id] AND i.index_id = p.index_id
)
END
,[total_space_MB] = cast(round(( au.total_pages * (8/1024.00)), 2) AS DECIMAL(36,2))
,[used_space_MB] = cast(round(( au.used_pages * (8/1024.00)), 2) AS DECIMAL(36,2))
,[unused_space_MB] = cast(round(((au.total_pages - au.used_pages) * (8/1024.00)), 2) AS DECIMAL(36,2))
FROM sys.schemas s
JOIN sys.tables t ON s.schema_id = t.schema_id
JOIN sys.indexes i ON t.object_id = i.object_id
JOIN (
SELECT [object_id], index_id, partition_count=count(*), [rows]=sum([rows]), data_compression_cnt=count(distinct [data_compression])
FROM sys.partitions
GROUP BY [object_id], [index_id]
) p ON i.[object_id] = p.[object_id] AND i.[index_id] = p.[index_id]
JOIN (
SELECT p.[object_id], p.[index_id], total_pages = sum(a.total_pages), used_pages = sum(a.used_pages), data_pages=sum(a.data_pages)
FROM sys.partitions p
JOIN sys.allocation_units a ON p.[partition_id] = a.[container_id]
GROUP BY p.[object_id], p.[index_id]
) au ON i.[object_id] = au.[object_id] AND i.[index_id] = au.[index_id]
WHERE t.is_ms_shipped = 0 -- Not a system table
我要感谢Greg Low先生的提问:
SELECT o.name AS ObjectName,
SUM(reserved_page_count) * 8.0 / 1024 AS SizeinMB
FROM sys.dm_db_partition_stats AS ps
INNER JOIN sys.sysobjects AS o
ON ps.object_id = o.id
GROUP BY o.name
ORDER BY SizeinMB DESC;
与Marc_s的回答有一点不同,因为我经常回到这一页,按大多数第一行排序:
SELECT
t.NAME AS TableName,
s.Name AS SchemaName,
p.rows AS RowCounts,
SUM(a.total_pages) * 8 AS TotalSpaceKB,
SUM(a.used_pages) * 8 AS UsedSpaceKB,
(SUM(a.total_pages) - SUM(a.used_pages)) * 8 AS UnusedSpaceKB
FROM
sys.tables t
INNER JOIN
sys.indexes i ON t.OBJECT_ID = i.object_id
INNER JOIN
sys.partitions p ON i.object_id = p.OBJECT_ID AND i.index_id = p.index_id
INNER JOIN
sys.allocation_units a ON p.partition_id = a.container_id
LEFT OUTER JOIN
sys.schemas s ON t.schema_id = s.schema_id
WHERE
t.NAME NOT LIKE 'dt%'
AND t.is_ms_shipped = 0
AND i.OBJECT_ID > 255
GROUP BY
t.Name, s.Name, p.Rows
ORDER BY
--p.rows DESC --Uncomment to order by amount rows instead of size in KB.
SUM(a.total_pages) DESC