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

如果您只关心数据库中的空浪费空间,而不关心单个表,则可以考虑以下问题:

如果数据库经历了大量的数据插入和删除,可能与ETL情况类似,这将导致数据库中有太多未使用的空间,因为文件组会自动增长,但不会自动收缩。

您可以通过使用数据库的财产页面来查看是否是这种情况。您可以收缩(右键单击数据库>任务>收缩)并收回一些空间。但是,如果根本原因仍然存在,则数据库将增长(并花费额外的时间尝试增长,直到增长到足够的速度,所以不要这样做)

其他回答

下面是另一种方法:使用SQLServerManagementStudio,在对象资源管理器中,转到数据库并选择表

然后打开“对象浏览器详细信息”(按F7或转到“查看”->“对象浏览器详情”)。在对象资源管理器详细信息页面中,右键单击列标题并启用您希望在页面中看到的列。您也可以按任何列对数据进行排序。

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

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;

我发现这个查询很容易使用和快速。

select schema_name(tab.schema_id) + '.' + tab.name as [table], 
cast(sum(spc.used_pages * 8)/1024.00 as numeric(36, 2)) as used_mb,
cast(sum(spc.total_pages * 8)/1024.00 as numeric(36, 2)) as allocated_mb
from sys.tables (nolock) tab
inner join sys.indexes (nolock) ind 
    on tab.object_id = ind.object_id
inner join sys.partitions  (nolock) part 
    on ind.object_id = part.object_id and ind.index_id = part.index_id
inner join sys.allocation_units (nolock) spc
    on part.partition_id = spc.container_id
group by schema_name(tab.schema_id) + '.' + tab.name
order by sum(spc.used_pages) desc