如何在SQL Server数据库中查找最大的对象?

时间:2010-01-19 15:09:40

标签: sql-server database

如何在SQL Server数据库中查找最大的对象?首先,通过确定哪些表(和相关索引)最大,然后确定特定表中哪些行最大(我们将二进制数据存储在BLOB中)?

是否有任何工具可以帮助进行此类数据库分析?或者是否有一些我可以针对系统表运行的简单查询?

7 个答案:

答案 0 :(得分:254)

我一直在使用这个SQL脚本(我从某人那里得到的 - 无法重建它的来源)多年来它帮助我理解并确定了索引和表格的大小:

SELECT 
    t.name AS TableName,
    i.name as indexName,
    sum(p.rows) as RowCounts,
    sum(a.total_pages) as TotalPages, 
    sum(a.used_pages) as UsedPages, 
    sum(a.data_pages) as DataPages,
    (sum(a.total_pages) * 8) / 1024 as TotalSpaceMB, 
    (sum(a.used_pages) * 8) / 1024 as UsedSpaceMB, 
    (sum(a.data_pages) * 8) / 1024 as DataSpaceMB
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
WHERE 
    t.name NOT LIKE 'dt%' AND
    i.object_id > 255 AND  
    i.index_id <= 1
GROUP BY 
    t.name, i.object_id, i.index_id, i.name 
ORDER BY 
    object_name(i.object_id) 

当然,您可以使用其他订购标准,例如

ORDER BY SUM(p.rows) DESC

获取具有最多行的表,或

ORDER BY SUM(a.total_pages) DESC

获取使用最多页面(8K块)的表格。

答案 1 :(得分:56)

在SQL Server 2008中,您还可以按顶级表运行标准报告“磁盘使用情况”。这可以通过右键单击数据库,选择报告 - &gt;标准报告并选择所需报告来找到。

答案 2 :(得分:3)

您也可以使用以下代码:

USE AdventureWork
GO
CREATE TABLE #GetLargest 
(
  table_name    sysname ,
  row_count     INT,
  reserved_size VARCHAR(50),
  data_size     VARCHAR(50),
  index_size    VARCHAR(50),
  unused_size   VARCHAR(50)
)

SET NOCOUNT ON

INSERT #GetLargest

EXEC sp_msforeachtable 'sp_spaceused ''?'''

SELECT 
  a.table_name,
  a.row_count,
  COUNT(*) AS col_count,
  a.data_size
  FROM #GetLargest a
     INNER JOIN information_schema.columns b
     ON a.table_name collate database_default
     = b.table_name collate database_default
       GROUP BY a.table_name, a.row_count, a.data_size
       ORDER BY CAST(REPLACE(a.data_size, ' KB', '') AS integer) DESC

DROP TABLE #GetLargest

答案 3 :(得分:3)

此查询有助于查找您所连接的最大表。

SELECT  TOP 1 OBJECT_NAME(OBJECT_ID) TableName, st.row_count
FROM sys.dm_db_partition_stats st
WHERE index_id < 2
ORDER BY st.row_count DESC

答案 4 :(得分:2)

如果您使用的是Sql Server Management Studio 2008,则可以在对象资源管理器详细信息窗口中查看某些数据字段。只需浏览并选择表格文件夹即可。在详细信息视图中,您可以右键单击列标题并在“报告”中添加字段。如果您使用SSMS 2008快递,您的里程可能会有所不同。

答案 5 :(得分:2)

我发现此查询在SqlServerCentral中也非常有用,这里是原帖的链接

Sql Server largest tables

  select name=object_schema_name(object_id) + '.' + object_name(object_id)
, rows=sum(case when index_id < 2 then row_count else 0 end)
, reserved_kb=8*sum(reserved_page_count)
, data_kb=8*sum( case 
     when index_id<2 then in_row_data_page_count + lob_used_page_count + row_overflow_used_page_count 
     else lob_used_page_count + row_overflow_used_page_count 
    end )
, index_kb=8*(sum(used_page_count) 
    - sum( case 
           when index_id<2 then in_row_data_page_count + lob_used_page_count + row_overflow_used_page_count 
        else lob_used_page_count + row_overflow_used_page_count 
        end )
     )    
, unused_kb=8*sum(reserved_page_count-used_page_count)
from sys.dm_db_partition_stats
where object_id > 1024
group by object_id
order by 
rows desc   

在我的数据库中,他们在此查询和第一个答案之间给出了不同的结果。

希望有人找到有用的

答案 6 :(得分:0)

@marc_s的回答非常好,我已经使用了几年了。但是,我注意到该脚本错过了某些列存储索引中的数据,并且没有显示完整的图片。例如。当您对脚本执行SUM(TotalSpace)并将其与Management Studio中的总空间数据库属性进行比较时,在我的情况下数字不匹配(Management Studio显示较大的数字)。我修改了脚本以克服此问题并对其进行了扩展:

select
    tables.[name] as table_name,
    schemas.[name] as schema_name,
    isnull(db_name(dm_db_index_usage_stats.database_id), 'Unknown') as database_name,
    sum(allocation_units.total_pages) * 8 as total_space_kb,
    cast(round(((sum(allocation_units.total_pages) * 8) / 1024.00), 2) as numeric(36, 2)) as total_space_mb,
    sum(allocation_units.used_pages) * 8 as used_space_kb,
    cast(round(((sum(allocation_units.used_pages) * 8) / 1024.00), 2) as numeric(36, 2)) as used_space_mb,
    (sum(allocation_units.total_pages) - sum(allocation_units.used_pages)) * 8 as unused_space_kb,
    cast(round(((sum(allocation_units.total_pages) - sum(allocation_units.used_pages)) * 8) / 1024.00, 2) as numeric(36, 2)) as unused_space_mb,
    count(distinct indexes.index_id) as indexes_count,
    max(dm_db_partition_stats.row_count) as row_count,
    iif(max(isnull(user_seeks, 0)) = 0 and max(isnull(user_scans, 0)) = 0 and max(isnull(user_lookups, 0)) = 0, 1, 0) as no_reads,
    iif(max(isnull(user_updates, 0)) = 0, 1, 0) as no_writes,
    max(isnull(user_seeks, 0)) as user_seeks,
    max(isnull(user_scans, 0)) as user_scans,
    max(isnull(user_lookups, 0)) as user_lookups,
    max(isnull(user_updates, 0)) as user_updates,
    max(last_user_seek) as last_user_seek,
    max(last_user_scan) as last_user_scan,
    max(last_user_lookup) as last_user_lookup,
    max(last_user_update) as last_user_update,
    max(tables.create_date) as create_date,
    max(tables.modify_date) as modify_date
from 
    sys.tables
    left join sys.schemas on schemas.schema_id = tables.schema_id
    left join sys.indexes on tables.object_id = indexes.object_id
    left join sys.partitions on indexes.object_id = partitions.object_id and indexes.index_id = partitions.index_id
    left join sys.allocation_units on partitions.partition_id = allocation_units.container_id
    left join sys.dm_db_index_usage_stats on tables.object_id = dm_db_index_usage_stats.object_id and indexes.index_id = dm_db_index_usage_stats.index_id
    left join sys.dm_db_partition_stats on tables.object_id = dm_db_partition_stats.object_id and indexes.index_id = dm_db_partition_stats.index_id
group by schemas.[name], tables.[name], isnull(db_name(dm_db_index_usage_stats.database_id), 'Unknown')
order by 5 desc

希望对某人有帮助。 此脚本已针对具有数百种不同表,索引和架构的大型TB范围的数据库进行了测试。