How to find largest objects in a SQL Server database?

Question

How would I go about finding the largest objects in a SQL Server database? First, by determining which tables (and related indices) are the largest and then determining whi

Answer 1

This query help to find largest table in you are connection.

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

Answer 2

If you are using Sql Server Management Studio 2008 there are certain data fields you can view in the object explorer details window. Simply browse to and select the tables folder. In the details view you are able to right-click the column titles and add fields to the "report". Your mileage may vary if you are on SSMS 2008 express.

Answer 3

In SQL Server 2008, you can also just run the standard report Disk Usage by Top Tables. This can be found by right clicking the DB, selecting Reports->Standard Reports and selecting the report you want.

Answer 4

@marc_s's answer is very great and I've been using it for few years. However, I noticed that the script misses data in some columnstore indexes and doesn't show complete picture. E.g. when you do SUM(TotalSpace) against the script and compare it with total space database property in Management Studio the numbers don't match in my case (Management Studio shows larger numbers). I modified the script to overcome this issue and extended it a little bit:

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

Hope it will be helpful for someone. This script was tested against large TB-wide databases with hundreds of different tables, indexes and schemas.

Answer 5

I've been using this SQL script (which I got from someone, somewhere - can't reconstruct who it came from) for ages and it's helped me quite a bit understanding and determining the size of indices and tables:

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) 

Of course, you can use another ordering criteria, e.g.

ORDER BY SUM(p.rows) DESC

to get the tables with the most rows, or

ORDER BY SUM(a.total_pages) DESC

to get the tables with the most pages (8K blocks) used.

Answer 6

You may also use the following code:

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