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;

其他回答

这将为您提供每个表的大小和记录计数。

set ANSI_NULLS ON
set QUOTED_IDENTIFIER ON
GO
-- Get a list of tables and their sizes on disk
ALTER PROCEDURE [dbo].[sp_Table_Sizes]
AS
BEGIN
    -- SET NOCOUNT ON added to prevent extra result sets from
    -- interfering with SELECT statements.
    SET NOCOUNT ON;
DECLARE @table_name VARCHAR(500)  
DECLARE @schema_name VARCHAR(500)  
DECLARE @tab1 TABLE( 
        tablename VARCHAR (500) collate database_default 
       ,schemaname VARCHAR(500) collate database_default 
) 

CREATE TABLE #temp_Table ( 
        tablename sysname 
       ,row_count INT 
       ,reserved VARCHAR(50) collate database_default 
       ,data VARCHAR(50) collate database_default 
       ,index_size VARCHAR(50) collate database_default 
       ,unused VARCHAR(50) collate database_default  
) 

INSERT INTO @tab1  
SELECT Table_Name, Table_Schema  
FROM information_schema.tables  
WHERE TABLE_TYPE = 'BASE TABLE' 

DECLARE c1 CURSOR FOR 
SELECT Table_Schema + '.' + Table_Name   
FROM information_schema.tables t1  
WHERE TABLE_TYPE = 'BASE TABLE' 

OPEN c1 
FETCH NEXT FROM c1 INTO @table_name 
WHILE @@FETCH_STATUS = 0  
BEGIN   
        SET @table_name = REPLACE(@table_name, '[','');  
        SET @table_name = REPLACE(@table_name, ']','');  

        -- make sure the object exists before calling sp_spacedused 
        IF EXISTS(SELECT id FROM sysobjects WHERE id = OBJECT_ID(@table_name)) 
        BEGIN 
               INSERT INTO #temp_Table EXEC sp_spaceused @table_name, false; 
        END 

        FETCH NEXT FROM c1 INTO @table_name 
END 
CLOSE c1 
DEALLOCATE c1 

SELECT  t1.* 
       ,t2.schemaname  
FROM #temp_Table t1  
INNER JOIN @tab1 t2 ON (t1.tablename = t2.tablename ) 
ORDER BY schemaname,t1.tablename; 

DROP TABLE #temp_Table
END

我在marc_s答案的顶部添加了几列:

with fs
as
(
select i.object_id,
        p.rows AS RowCounts,
        SUM(a.total_pages) * 8 AS TotalSpaceKb
from     sys.indexes i 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
WHERE 
    i.OBJECT_ID > 255 
GROUP BY 
    i.object_id,
    p.rows
)

SELECT 
    t.NAME AS TableName,
    fs.RowCounts,
    fs.TotalSpaceKb,
    t.create_date,
    t.modify_date,
    ( select COUNT(1)
        from sys.columns c 
        where c.object_id = t.object_id ) TotalColumns    
FROM 
    sys.tables t INNER JOIN      
    fs  ON t.OBJECT_ID = fs.object_id
WHERE 
    t.NAME NOT LIKE 'dt%' 
    AND t.is_ms_shipped = 0
ORDER BY 
    t.Name

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。

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

作为marc_s答案(已被接受的答案)的一个简单扩展,它被调整为返回列计数并允许过滤:

SELECT *
FROM
(

SELECT 
    t.NAME AS TableName,
    s.Name AS SchemaName,
    p.rows AS RowCounts,
    COUNT(DISTINCT c.COLUMN_NAME) as ColumnCount,
    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
INNER JOIN
    INFORMATION_SCHEMA.COLUMNS c ON t.NAME = c.TABLE_NAME
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
) AS Result

WHERE
    RowCounts > 1000
    AND ColumnCount > 10
ORDER BY 
    UsedSpaceKB DESC