Sql как найти самую большую таблицу

Дек 15, 2014 21:48

@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.

From time to time, I need to find the largest tables in terms of rows / used space / total space etc. I originally used sp_spaceused, a temp table and a cursor (or sp_MSForEachTable) and put them into a stored procedure to get the information I need.

But now as I am working more on table partition management, I find there is a DMV that can give me the exactly info I need, and what surprises me is that sp_spaceused is actually using this DMV, i.e. sys.dm_db_partition_stats to get the data as well. So here is the one sql to get the largest tables in a db (only applicable to sql server 2k5 / 2k8 for tables without xml indexes)

— find the table size info (no xml index) using sys.dm_db_partition_stats
— Author: Jeffrey Yao
— Date: Sept 27, 2010

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
— data_kb desc
— reserved_kb desc
— data_kb desc
— index_kb desc
— unsed_kb desc