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

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

其他回答

SELECT 
    t.NAME AS TableName,
    s.Name AS SchemaName,
    p.rows,
    SUM(a.total_pages) * 8 AS TotalSpaceKB, 
    CAST(ROUND(((SUM(a.total_pages) * 8) / 1024.00), 2) AS NUMERIC(36, 2)) AS TotalSpaceMB,
    SUM(a.used_pages) * 8 AS UsedSpaceKB, 
    CAST(ROUND(((SUM(a.used_pages) * 8) / 1024.00), 2) AS NUMERIC(36, 2)) AS UsedSpaceMB, 
    (SUM(a.total_pages) - SUM(a.used_pages)) * 8 AS UnusedSpaceKB,
    CAST(ROUND(((SUM(a.total_pages) - SUM(a.used_pages)) * 8) / 1024.00, 2) AS NUMERIC(36, 2)) AS UnusedSpaceMB
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 
    TotalSpaceMB DESC, t.Name

我们使用表分区,由于重复记录,上面提供的查询有一些问题。

对于需要此功能的人,您可以在下面找到SQL Server 2014在生成“磁盘使用情况(按表)”报告时运行的查询。我假设它也适用于以前版本的SQL Server。

它就像一个符咒。

SELECT
    a2.name AS [tablename],
    a1.rows as row_count,
    (a1.reserved + ISNULL(a4.reserved,0))* 8 AS reserved, 
    a1.data * 8 AS data,
    (CASE WHEN (a1.used + ISNULL(a4.used,0)) > a1.data THEN (a1.used + ISNULL(a4.used,0)) - a1.data ELSE 0 END) * 8 AS index_size,
    (CASE WHEN (a1.reserved + ISNULL(a4.reserved,0)) > a1.used THEN (a1.reserved + ISNULL(a4.reserved,0)) - a1.used ELSE 0 END) * 8 AS unused
FROM
    (SELECT 
        ps.object_id,
        SUM (
            CASE
                WHEN (ps.index_id < 2) THEN row_count
                ELSE 0
            END
            ) AS [rows],
        SUM (ps.reserved_page_count) AS reserved,
        SUM (
            CASE
                WHEN (ps.index_id < 2) THEN (ps.in_row_data_page_count + ps.lob_used_page_count + ps.row_overflow_used_page_count)
                ELSE (ps.lob_used_page_count + ps.row_overflow_used_page_count)
            END
            ) AS data,
        SUM (ps.used_page_count) AS used
    FROM sys.dm_db_partition_stats ps
        WHERE ps.object_id NOT IN (SELECT object_id FROM sys.tables WHERE is_memory_optimized = 1)
    GROUP BY ps.object_id) AS a1
LEFT OUTER JOIN 
    (SELECT 
        it.parent_id,
        SUM(ps.reserved_page_count) AS reserved,
        SUM(ps.used_page_count) AS used
     FROM sys.dm_db_partition_stats ps
     INNER JOIN sys.internal_tables it ON (it.object_id = ps.object_id)
     WHERE it.internal_type IN (202,204)
     GROUP BY it.parent_id) AS a4 ON (a4.parent_id = a1.object_id)
INNER JOIN sys.all_objects a2  ON ( a1.object_id = a2.object_id ) 
INNER JOIN sys.schemas a3 ON (a2.schema_id = a3.schema_id)
WHERE a2.type <> N'S' and a2.type <> N'IT'
ORDER BY a3.name, a2.name

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)

CREATE TABLE #tmp_table_info
(
id int identity(1,1),
tblname varchar(200)
);
CREATE TABLE #SpaceUsed 
(
     TableName sysname
    ,NumRows BIGINT
    ,ReservedSpace VARCHAR(50)
    ,DataSpace VARCHAR(50)
    ,IndexSize VARCHAR(50)
    ,UnusedSpace VARCHAR(50)
) 

insert into #tmp_table_info
select s.name+'.'+t.name 
from sys.tables t 
inner join sys.schemas s on t.schema_id = s.schema_id
where t.type = 'U';

declare @min int =1,@max int = 0
select @max = count(*)
from #tmp_table_info
while(@min<=@max)
begin
    declare @tablename varchar(200)
    select @tablename=tblname
    from #tmp_table_info
    where id =@min

    DECLARE @str VARCHAR(500)
    SET @str =  'sp_spaceused '''+@tablename+''''
    INSERT INTO #SpaceUsed 
    EXEC (@str)  
    set @min =@min + 1
end;
select @@SERVERNAME as servername,DB_NAME() as DatabaseName,CONVERT(numeric(18,0),REPLACE(ReservedSpace,' KB','')) / 1024 as ReservedSpace_MB,
CONVERT(numeric(18,0),REPLACE(DataSpace,' KB','')) / 1024 as DataSpace_MB,
CONVERT(numeric(18,0),REPLACE(IndexSize,' KB','')) / 1024 as IndexSpace_MB,
CONVERT(numeric(18,0),REPLACE(UnusedSpace,' KB','')) / 1024 as UnusedSpace_MB from #SpaceUsed
drop table #tmp_table_info
drop table #SpaceUsed

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