Comments on: Estimating the Size of a Table in SQL Server 2008

By: Ishwar Nataraj

Ishwar Nataraj — Tue, 10 Jan 2012 04:32:10 +0000

sp_spaceused or sp_helpdb is also a good way to measure table growth and you can basically determie growth levels based in an exponential manner.

By: @billinkc

@billinkc — Thu, 11 Aug 2011 17:19:59 +0000

Thanks for the script, assuming I’m not missing something, I think you can simplify the population of generation of rows for the insert into @tableColumn with the following query

DECLARE @TableColumn TableColumnDefinition

DECLARE
@table_name sysname
, @schema_name sysname
SELECT
@table_name = ‘TerribleTable’
, @schema_name = ‘mySchema’

; WITH TABLE_COLUMNS AS
(
SELECT
T.name AS [DataType]
, SC.max_length AS [Length]
, SC.name As column_name
, ST.name AS table_name
, schema_name(ST.schema_id) AS [schema_name]
, ST.object_id
FROM
sys.tables ST
INNER JOIN
sys.columns SC
ON SC.object_id = ST.object_id
INNER JOIN
sys.types T
ON T.system_type_id = SC.system_type_id
)
, PRIMARY_KEYS AS
(
SELECT DISTINCT
I.object_id
, IC.column_id
, I.fill_factor
, I.is_primary_key
, C.name AS column_name
FROM
sys.indexes I
INNER JOIN
sys.index_columns IC
ON IC.object_id = I.object_id
AND IC.index_id = I.index_id
INNER JOIN
sys.columns C
ON C.object_id = I.object_id
AND C.column_id = IC.column_id
WHERE
I.is_primary_key = 1
)
INSERT INTO @TableColumn
SELECT
TC.[DataType]
, TC.[Length]
, PK.is_primary_key AS [IsKey]
FROM
TABLE_COLUMNS TC
LEFT OUTER JOIN
PRIMARY_KEYS PK
ON PK.object_id = TC.object_id
AND PK.column_name = TC.column_name
WHERE
TC.table_name = @table_name
AND TC.[schema_name] = @schema_name

By: selcuk

selcuk — Wed, 27 Apr 2011 22:46:53 +0000

Thanks….

By: Walter

Walter — Wed, 09 Jun 2010 19:05:25 +0000

This is great Tim. Thanks for making this calculation so much easier.

By: Tim

Tim — Thu, 19 Nov 2009 20:27:56 +0000

Yes, you have to run it four times (once for each index).

The Clustered Index needs all the columns specified, because the clustered index basically IS the table (that’s why you can’t have “included columns” in clustered indexes). As for the other nonclustered indexes, you only need to specify one key column for each – if you don’t have any included columns in the nonclustered indexes, you don’t need to have anything other than the key column(s). Also note for your non-unique nonclustered indexes, set IsClusteredIndexUnique to 0, it should really be called “IsIndexUnique”.

We’re not really calculating the size of a table per-se because a table is simply the sum of the size of its indexes (let’s not talk about heaps – there’s no index there, but it certainly takes up space) – we’re just calculating the size of each index.

By: Oshik

Oshik — Thu, 19 Nov 2009 20:15:51 +0000

Hi Tim,
i think i miss something.
let me clear my request:
i have a partition table with 600 million rows, with 4 indexes.
one is a clustered index(three key columns)
secound – non-clustered index(non uniqe, one column)
third- non-clustered index(non uniqe, one column)
fourth- non-clustered index(non uniqe, one column).
my table has a 20 columns(varchar, int, datetime etc…).
as you meen, i need to create a TYPE(and setting the isKey to 1 for eash column index) with my table definition and run it foure time?
Thanks,
oshik.

By: Tim

Tim — Mon, 16 Nov 2009 22:42:38 +0000

For the multiple indexes, estimate each one individually. For non-clustered indexes, I just pretend it’s a table and included columns are non-key columns – it seems to come out pretty close in my tests. (You can just test it out against existing indexes). As for partitioned tables and partition-aligned indexes, I’ve never done anything special to account for the fact that they’re partitioned. As a matter of fact, all the examples I mentioned above were indeed partitioned tables and partition-aligned indexes.

By: Oshik

Oshik — Mon, 16 Nov 2009 13:50:45 +0000

hi TIM,
how the stored procedure Utility_EstimateClusteredIndexSize know how muth indexes i have in specific table?
how it’s know to calculate table with six indexes(one clustered and five nonclustered index)
what about partition table?
Thanks,
Oshik.

By: Tim

Tim — Thu, 12 Nov 2009 19:30:52 +0000

Sure, if you can let me know what parameters you’re passing to the procedure, I can take a look and try to figure out why it’s not working as expected

By: vinayak

vinayak — Thu, 12 Nov 2009 11:02:40 +0000

Hi,

i tried to run this. Its giving so many errors.first i executed TableColumnDefinition and theni executed the procedure.its thorowing so many exceptions. can u pls tel the way to run this?

Thanks in advance.