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	<title>Comments on: SQL SERVER &#8211; 2005 &#8211; Database Table Partitioning Tutorial &#8211; How to Horizontal Partition Database Table</title>
	<atom:link href="http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/feed/" rel="self" type="application/rss+xml" />
	<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/</link>
	<description>Personal Notes of Pinal Dave</description>
	<lastBuildDate>Thu, 09 Feb 2012 19:36:10 +0000</lastBuildDate>
	<sy:updatePeriod>hourly</sy:updatePeriod>
	<sy:updateFrequency>1</sy:updateFrequency>
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	<item>
		<title>By: SQL SERVER &#8211; Advantages of Partitioning &#8211; Quiz &#8211; Puzzle &#8211; 30 of 31 &#171; SQL Server Journey with SQL Authority</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-245879</link>
		<dc:creator><![CDATA[SQL SERVER &#8211; Advantages of Partitioning &#8211; Quiz &#8211; Puzzle &#8211; 30 of 31 &#171; SQL Server Journey with SQL Authority]]></dc:creator>
		<pubDate>Tue, 31 Jan 2012 01:31:56 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-245879</guid>
		<description><![CDATA[[...] Server Interview Questions and Answers ISBN: 1466405643 Page#168-170 Database Table Partitioning Tutorial – How to Horizontal Partition Database Table White Paper – Partitioned Table and Index Strategies Using SQL Server [...]]]></description>
		<content:encoded><![CDATA[<p>[...] Server Interview Questions and Answers ISBN: 1466405643 Page#168-170 Database Table Partitioning Tutorial – How to Horizontal Partition Database Table White Paper – Partitioned Table and Index Strategies Using SQL Server [...]</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Sivakumar</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-226803</link>
		<dc:creator><![CDATA[Sivakumar]]></dc:creator>
		<pubDate>Tue, 27 Dec 2011 15:06:02 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-226803</guid>
		<description><![CDATA[It is very helpful and useful guide for beginners.  Excellent Job.

Thanks Pinal. Keep it up.]]></description>
		<content:encoded><![CDATA[<p>It is very helpful and useful guide for beginners.  Excellent Job.</p>
<p>Thanks Pinal. Keep it up.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Manish</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-222868</link>
		<dc:creator><![CDATA[Manish]]></dc:creator>
		<pubDate>Wed, 21 Dec 2011 12:09:25 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-222868</guid>
		<description><![CDATA[I got the answer of my question, 

Select $partition.TESTDB_PARTITIONRANGE(id) AS [Partition Number],* from testtable 
where $partition.TESTDB_PARTITIONRANGE(id)=1

if some one is having any other or better way, please share.


Thanks
Manish]]></description>
		<content:encoded><![CDATA[<p>I got the answer of my question, </p>
<p>Select $partition.TESTDB_PARTITIONRANGE(id) AS [Partition Number],* from testtable<br />
where $partition.TESTDB_PARTITIONRANGE(id)=1</p>
<p>if some one is having any other or better way, please share.</p>
<p>Thanks<br />
Manish</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Manish</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-221913</link>
		<dc:creator><![CDATA[Manish]]></dc:creator>
		<pubDate>Tue, 20 Dec 2011 07:36:34 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-221913</guid>
		<description><![CDATA[Hi every one, 

How to view the data of any specific partition ?

Thanks
Manish]]></description>
		<content:encoded><![CDATA[<p>Hi every one, </p>
<p>How to view the data of any specific partition ?</p>
<p>Thanks<br />
Manish</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Sourabh Jain</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-211239</link>
		<dc:creator><![CDATA[Sourabh Jain]]></dc:creator>
		<pubDate>Mon, 05 Dec 2011 11:02:01 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-211239</guid>
		<description><![CDATA[Once the partitions are created on the tables , the information is stored in the  SQL Server metadata tables. 
Now when the select queries are fired , the metadata information is compared  with the where clause and appropriate partitions are only scanned for retrieval of data and not all the partitions.

Think of it like this.. India is made up of 28 states... Relate each of the state as partition... Now if I need information of a person residing in Gujarat , the query will hit only the people information staying in gujarat as opposed to all 28 states and thereby improving performance of the query]]></description>
		<content:encoded><![CDATA[<p>Once the partitions are created on the tables , the information is stored in the  SQL Server metadata tables.<br />
Now when the select queries are fired , the metadata information is compared  with the where clause and appropriate partitions are only scanned for retrieval of data and not all the partitions.</p>
<p>Think of it like this.. India is made up of 28 states&#8230; Relate each of the state as partition&#8230; Now if I need information of a person residing in Gujarat , the query will hit only the people information staying in gujarat as opposed to all 28 states and thereby improving performance of the query</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Sourabh Jain</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-211235</link>
		<dc:creator><![CDATA[Sourabh Jain]]></dc:creator>
		<pubDate>Mon, 05 Dec 2011 10:54:22 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-211235</guid>
		<description><![CDATA[This is being used in order to check the existence of database. If the TestDB exists , the same is being dropped.]]></description>
		<content:encoded><![CDATA[<p>This is being used in order to check the existence of database. If the TestDB exists , the same is being dropped.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Sourabh Jain</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-211232</link>
		<dc:creator><![CDATA[Sourabh Jain]]></dc:creator>
		<pubDate>Mon, 05 Dec 2011 10:52:39 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-211232</guid>
		<description><![CDATA[Yes Kenneth... In order to make use of partitions effectively, the query&#039;s where clause should have the column information associated to partitions.

Since the query wont have the account information , it will have to check all partitions to get the data assuming the partitions are created on Account]]></description>
		<content:encoded><![CDATA[<p>Yes Kenneth&#8230; In order to make use of partitions effectively, the query&#8217;s where clause should have the column information associated to partitions.</p>
<p>Since the query wont have the account information , it will have to check all partitions to get the data assuming the partitions are created on Account</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Priyavadhani.S</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-207962</link>
		<dc:creator><![CDATA[Priyavadhani.S]]></dc:creator>
		<pubDate>Wed, 30 Nov 2011 13:13:57 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-207962</guid>
		<description><![CDATA[IF EXISTS (
SELECT name
FROM sys.databases
WHERE name = N&#039;TestDB&#039;)
DROP DATABASE TestDB;
for what purpose we are using it 
kindly tell me]]></description>
		<content:encoded><![CDATA[<p>IF EXISTS (<br />
SELECT name<br />
FROM sys.databases<br />
WHERE name = N&#8217;TestDB&#8217;)<br />
DROP DATABASE TestDB;<br />
for what purpose we are using it<br />
kindly tell me</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: sanjay</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-205057</link>
		<dc:creator><![CDATA[sanjay]]></dc:creator>
		<pubDate>Sat, 26 Nov 2011 16:50:20 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-205057</guid>
		<description><![CDATA[very useful .thanks]]></description>
		<content:encoded><![CDATA[<p>very useful .thanks</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: arjun</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-190839</link>
		<dc:creator><![CDATA[arjun]]></dc:creator>
		<pubDate>Tue, 08 Nov 2011 12:47:13 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-190839</guid>
		<description><![CDATA[hi pinal,

partition column and UNIQUE clustered index key column is different. that time i created partioned for table. but table–&gt;properties–&gt;storage doesn’t show partition. but if i use select * from sys.partitions where object_name(object_id)=’tablename’. it shows partioning except clustered index..what is the difference?..is this correct way to partition?

advance  THANKS for you help]]></description>
		<content:encoded><![CDATA[<p>hi pinal,</p>
<p>partition column and UNIQUE clustered index key column is different. that time i created partioned for table. but table–&gt;properties–&gt;storage doesn’t show partition. but if i use select * from sys.partitions where object_name(object_id)=’tablename’. it shows partioning except clustered index..what is the difference?..is this correct way to partition?</p>
<p>advance  THANKS for you help</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: arjun</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-190816</link>
		<dc:creator><![CDATA[arjun]]></dc:creator>
		<pubDate>Tue, 08 Nov 2011 12:09:01 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-190816</guid>
		<description><![CDATA[hi pinal,

partition column and clustered index key column is different. that time i created partioned for table. but table--&gt;properties--&gt;storage doesn&#039;t show partition. but if i use select * from sys.partitions where object_name(object_id)=&#039;tablename&#039;. it show partioning except clustered index..what is the difference? 

Thanks in advance for you help]]></description>
		<content:encoded><![CDATA[<p>hi pinal,</p>
<p>partition column and clustered index key column is different. that time i created partioned for table. but table&#8211;&gt;properties&#8211;&gt;storage doesn&#8217;t show partition. but if i use select * from sys.partitions where object_name(object_id)=&#8217;tablename&#8217;. it show partioning except clustered index..what is the difference? </p>
<p>Thanks in advance for you help</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Anandan Kanagarajan</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-178868</link>
		<dc:creator><![CDATA[Anandan Kanagarajan]]></dc:creator>
		<pubDate>Fri, 14 Oct 2011 15:27:53 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-178868</guid>
		<description><![CDATA[Very nice tutorial about creating Partition in SQL Server. Could you please explain how to check the existing partition&#039;s various details like on which column the partition has been created,etc on a database  and how to remove the partition that has been created.]]></description>
		<content:encoded><![CDATA[<p>Very nice tutorial about creating Partition in SQL Server. Could you please explain how to check the existing partition&#8217;s various details like on which column the partition has been created,etc on a database  and how to remove the partition that has been created.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: sivanandan</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-176158</link>
		<dc:creator><![CDATA[sivanandan]]></dc:creator>
		<pubDate>Fri, 07 Oct 2011 17:34:04 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-176158</guid>
		<description><![CDATA[Hi Pinal,

How select command will work after partitioning ,while  retrieve the data from the table?]]></description>
		<content:encoded><![CDATA[<p>Hi Pinal,</p>
<p>How select command will work after partitioning ,while  retrieve the data from the table?</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Rejith Egodahewa</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-174589</link>
		<dc:creator><![CDATA[Rejith Egodahewa]]></dc:creator>
		<pubDate>Mon, 03 Oct 2011 11:14:34 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-174589</guid>
		<description><![CDATA[we&#039;ve a massive SQL Server 2005 DB(Enterprise) comprising nearly 1500 tables which is partitioned. I ported the same database (NV_25_0.mdf) to my PC having (Standard) similar version.  

But when attaching it has thrown an error saying that due to a partitioning this cannt be attached. 

Pls. advise me to remove the partition and attach it to SQL Server in my PC....]]></description>
		<content:encoded><![CDATA[<p>we&#8217;ve a massive SQL Server 2005 DB(Enterprise) comprising nearly 1500 tables which is partitioned. I ported the same database (NV_25_0.mdf) to my PC having (Standard) similar version.  </p>
<p>But when attaching it has thrown an error saying that due to a partitioning this cannt be attached. </p>
<p>Pls. advise me to remove the partition and attach it to SQL Server in my PC&#8230;.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Yousef ekhtiari</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-146079</link>
		<dc:creator><![CDATA[Yousef ekhtiari]]></dc:creator>
		<pubDate>Tue, 05 Jul 2011 07:13:00 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-146079</guid>
		<description><![CDATA[how big is your table?is your  partitioning data based on a time if so do use Right range]]></description>
		<content:encoded><![CDATA[<p>how big is your table?is your  partitioning data based on a time if so do use Right range</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Yousef ekhtiari</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-146077</link>
		<dc:creator><![CDATA[Yousef ekhtiari]]></dc:creator>
		<pubDate>Tue, 05 Jul 2011 07:07:42 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-146077</guid>
		<description><![CDATA[Dealing with massive table is not easy,recently I  had  to correct a partition scheme which contains 5*12  months&#039;  partitions .you should be very careful at the time of creating partition scheme and function since a minor mistake might cause a serious problem that correcting  it may seems impossible as my case.
My advice to all is to understand its concept  completely.]]></description>
		<content:encoded><![CDATA[<p>Dealing with massive table is not easy,recently I  had  to correct a partition scheme which contains 5*12  months&#8217;  partitions .you should be very careful at the time of creating partition scheme and function since a minor mistake might cause a serious problem that correcting  it may seems impossible as my case.<br />
My advice to all is to understand its concept  completely.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Sudheshna</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-144646</link>
		<dc:creator><![CDATA[Sudheshna]]></dc:creator>
		<pubDate>Wed, 29 Jun 2011 12:03:14 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-144646</guid>
		<description><![CDATA[Hi,

Thanks for such a nice article, I have a question regarding the partition, that is I thought it will divide the rows between two parts through out the insertion process but it has inserted only 10 rows in first part and all other in to part2, is there a way I can equally distribute the rows among two partitions.

Thanks &amp; Regards]]></description>
		<content:encoded><![CDATA[<p>Hi,</p>
<p>Thanks for such a nice article, I have a question regarding the partition, that is I thought it will divide the rows between two parts through out the insertion process but it has inserted only 10 rows in first part and all other in to part2, is there a way I can equally distribute the rows among two partitions.</p>
<p>Thanks &amp; Regards</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: خدمات آموزش مجازي نت آزمون</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-143191</link>
		<dc:creator><![CDATA[خدمات آموزش مجازي نت آزمون]]></dc:creator>
		<pubDate>Thu, 23 Jun 2011 12:28:20 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-143191</guid>
		<description><![CDATA[hi
how could i create a partition function.]]></description>
		<content:encoded><![CDATA[<p>hi<br />
how could i create a partition function.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Ritesh</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-132513</link>
		<dc:creator><![CDATA[Ritesh]]></dc:creator>
		<pubDate>Mon, 02 May 2011 13:30:42 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-132513</guid>
		<description><![CDATA[Here is the complete solution:

-------------------------------------------------------------------------------------------------

-- CREATE DATABASE
CREATE DATABASE TUTORIAL
GO

-------------------------------------------------------------------------------------------------

USE TUTORIAL
GO
CREATE PROCEDURE [dbo].[sp_dim_date_monthly_partition] @nRunDate datetime  = NULL
AS
BEGIN

	SET NOCOUNT ON
	SET @nRunDate = &#039;2011-05-02&#039;
	-- Constants used in this procedure
	DECLARE @cRtnSuccess	integer
	SET		@cRtnSuccess	= 0

	DECLARE @cRtnFailed		integer
	SET		@cRtnFailed		= 1

	-- Setting up variables used in this procedure
	DECLARE @ErrCount		int
	SET		@ErrCount		= 0
	
	-- Obtaining the 1st at midnight of given date, if no date is given use the current date
	DECLARE @nDate			datetime
	DECLARE @dateid			numeric(8,0)
	SET		@nDate			= DATEADD(mm, DATEDIFF(mm,0,ISNULL(@nRunDate,GetDate())), 0)
    SELECT	@dateid			= (YEAR(@nDate) * 10000) + (MONTH(@nDate) * 100) + DAY(@nDate)

	-- Declare Database pathname variables
	DECLARE @db_name		nvarchar(4000)
	DECLARE @path			nvarchar(max)
	
	-- Declare Filegoup variables
	DECLARE @fg_name		nvarchar(max)
	DECLARE @filegroup_sql	nvarchar(max)

	-- Declare File variables
	DECLARE @file_name		nvarchar(max)
	DECLARE @file_sql		nvarchar(max)

	-- Declare Partition Variables
	DECLARE @ps_name		nvarchar(max)
	DECLARE @pf_name		nvarchar(max)
	DECLARE @pf_sql			nvarchar(max)
	DECLARE @ps_sql			nvarchar(max)

	-- Set Variable values
	SET		@db_name		= &#039;Tutorial&#039;
	SET		@fg_name		= &#039;fg_date_monthly&#039;
	SET		@file_name		= &#039;saledate_&#039; + replace(convert(varchar(7),@nDate,121),&#039;-&#039;,&#039;&#039;) 
	SET		@ps_name		= &#039;ps_date_monthly&#039;
	SET		@pf_name		= &#039;pf_date_monthly&#039;

	-- DETERMINE DATABASE DATA PATHNAME
	SELECT	@path = [filename] FROM master..sysdatabases WHERE NAME like @db_name
	IF charindex(&#039;\&#039;, @path) &gt; 0 
	BEGIN
		SELECT @path = substring(@path, 1, len(@path) - charindex(&#039;\&#039;, reverse(@path)))
	END
	SET		@path = @path

-------------------------------------------------------------------------------------------------
	
	-- ADD FILEGROUP
	IF Not Exists (SELECT * FROM sys.filegroups WHERE NAME = @fg_name)
	SET		@filegroup_sql =	
		&#039;ALTER DATABASE &#039; + QuoteName(@db_name,&#039;&#039;) + &#039; &#039; + 	&#039;ADD FILEGROUP &#039; + QuoteName(@fg_name,&#039;&#039;)
	EXEC( @filegroup_sql )

-------------------------------------------------------------------------------------------------
	
	-- ADD FILE TO FILEGROUP
	IF NOT EXISTS (SELECT * FROM sys.sysfiles WHERE NAME = @file_name)
	SET		@file_sql = 
			&#039;ALTER DATABASE&#039; + &#039; &#039; + QuoteName(@db_name,&#039;&#039;) + &#039; &#039; + &#039;
			 ADD FILE(	Name = &#039; + QuoteName(@file_name,&#039;&#039;&#039;&#039;) + &#039;, 
						FileName = &#039; + QuoteName( @Path + &#039;\&#039; + @file_name + &#039;.ndf&#039;, &#039;&#039;&#039;&#039; ) + &#039;,
						SIZE = 3072 KB, MAXSIZE = UNLIMITED, FILEGROWTH = 1024 KB
					  ) TO FILEGROUP &#039; + @fg_name 
	EXEC( @file_sql )

-------------------------------------------------------------------------------------------------

	-- CREATE PARTITION FUNCTION
	IF NOT EXISTS(SELECT * FROM sys.partition_functions WHERE NAME  = @pf_name)
	SET		@pf_sql = 
			&#039;CREATE PARTITION FUNCTION &#039; +  QuoteName(@pf_name,&#039;&#039;) + &#039;(numeric(8,0)) &#039; +
			&#039;AS RANGE RIGHT FOR VALUES (20110301,20110401)&#039;
	EXEC( @pf_sql )

-------------------------------------------------------------------------------------------------

	-- CREATE PARTITION SCHEME
	IF NOT EXISTS(SELECT * FROM sys.partition_schemes WHERE NAME = @ps_name)
	SET		@ps_sql = 
			&#039;CREATE PARTITION SCHEME &#039; + QuoteName(@ps_name,&#039;&#039;) + &#039; &#039; +
			&#039;AS  PARTITION &#039; +  QuoteName(@pf_name,&#039;&#039;) + &#039; ALL TO (&#039; + @fg_name  + &#039;)&#039;
	EXEC( @ps_sql )
 
-------------------------------------------------------------------------------------------------

	DECLARE @table_sql	nvarchar(max)
	-- CREATE A TABLE tblsaledate TO USE PARTITION SCHEME
	IF NOT EXISTS(SELECT * FROM sys.objects WHERE NAME = &#039;tblsaledate&#039; AND type = &#039;U&#039;)
	SET		@table_sql = 
			&#039;CREATE TABLE dbo.tblsaledate
			(
			 sale_id int not null identity(1,1),
			 sale_date datetime not null,
			 sale_date_id	numeric(8,0)
			) ON &#039; + QuoteName(@ps_name,&#039;&#039;) + &#039;(sale_date_id)&#039;
	EXEC( @table_sql )

	-- POPULATE tblsaledate TABLE with data
	DECLARE @startdate		datetime
	DECLARE @enddate		datetime
	DECLARE @thisdate		datetime
	DECLARE @day_id			int
	SET		@startdate		= &#039;2011-03-01&#039;
	SET		@enddate		= &#039;2011-06-01&#039;
	SET		@thisdate		= @startdate
	SET		@day_id			= 1
	WHILE	(@thisdate &lt;= @enddate)
	BEGIN
		INSERT INTO dbo.tblsaledate (sale_date, sale_date_id)
		VALUES (@thisdate, CAST(convert(varchar(8),@thisdate,112) As int))
		SET @day_id			= @day_id + 1
		SET @thisdate		= DATEADD(day, 1, @thisdate)
	END

---------------------------------------------------------------------------------------------------

	-- SET NEW FILE TO BE USED FOR NEXT SET OF DATA
	DECLARE	@aps_sql		nvarchar(max)
	SET		@aps_sql = 
			&#039;ALTER PARTITION SCHEME &#039; + QuoteName(@ps_name,&#039;&#039;) + &#039; NEXT USED &#039; + @fg_name
	EXEC( @aps_sql )

	-- ADD NEW RANGE FOR PARTITION
	DECLARE @apf_sql		nvarchar(max)
	SET		@apf_sql = 
			&#039;ALTER PARTITION FUNCTION &#039; + QuoteName(@pf_name,&#039;&#039;) + &#039;() 
			 SPLIT RANGE &#039; + &#039;(&#039; + CAST(@dateid as nvarchar) + &#039;)&#039;
	EXEC( @apf_sql )

---------------------------------------------------------------------------------------------------

	-- IMPLEMENT ERROR HANDLING
	SET		@ErrCount = @@ERROR
	IF		(@ErrCount  0)
	BEGIN
		RETURN @cRtnFailed
	END   
	SET NOCOUNT OFF;
	RETURN	@cRtnSuccess

-----------------------------------------------------------------------------------------------
/*

1. Data and Partition Views
============================

	----
	SELECT * , $partition.pf_date_monthly(sale_date_id) FROM dbo.tblsaledate

	----
	SELECT * FROM sys.partitions WHERE object_id = object_id(&#039;tblsaledate&#039;)

2. Usage
===================================
	----
	Exec [dbo].[sp_dim_date_monthly_partition]

	----
	DROP PROCEDURE [dbo].[sp_dim_date_monthly_partition]
	DROP PARTITION SCHEME [ps_date_monthly]
	DROP PARTITION FUNCTION [pf_date_monthly]

*/
---------------------------------------------------------------------------------------------------

END]]></description>
		<content:encoded><![CDATA[<p>Here is the complete solution:</p>
<p>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-</p>
<p>&#8211; CREATE DATABASE<br />
CREATE DATABASE TUTORIAL<br />
GO</p>
<p>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-</p>
<p>USE TUTORIAL<br />
GO<br />
CREATE PROCEDURE [dbo].[sp_dim_date_monthly_partition] @nRunDate datetime  = NULL<br />
AS<br />
BEGIN</p>
<p>	SET NOCOUNT ON<br />
	SET @nRunDate = &#8217;2011-05-02&#8242;<br />
	&#8211; Constants used in this procedure<br />
	DECLARE @cRtnSuccess	integer<br />
	SET		@cRtnSuccess	= 0</p>
<p>	DECLARE @cRtnFailed		integer<br />
	SET		@cRtnFailed		= 1</p>
<p>	&#8211; Setting up variables used in this procedure<br />
	DECLARE @ErrCount		int<br />
	SET		@ErrCount		= 0</p>
<p>	&#8211; Obtaining the 1st at midnight of given date, if no date is given use the current date<br />
	DECLARE @nDate			datetime<br />
	DECLARE @dateid			numeric(8,0)<br />
	SET		@nDate			= DATEADD(mm, DATEDIFF(mm,0,ISNULL(@nRunDate,GetDate())), 0)<br />
    SELECT	@dateid			= (YEAR(@nDate) * 10000) + (MONTH(@nDate) * 100) + DAY(@nDate)</p>
<p>	&#8211; Declare Database pathname variables<br />
	DECLARE @db_name		nvarchar(4000)<br />
	DECLARE @path			nvarchar(max)</p>
<p>	&#8211; Declare Filegoup variables<br />
	DECLARE @fg_name		nvarchar(max)<br />
	DECLARE @filegroup_sql	nvarchar(max)</p>
<p>	&#8211; Declare File variables<br />
	DECLARE @file_name		nvarchar(max)<br />
	DECLARE @file_sql		nvarchar(max)</p>
<p>	&#8211; Declare Partition Variables<br />
	DECLARE @ps_name		nvarchar(max)<br />
	DECLARE @pf_name		nvarchar(max)<br />
	DECLARE @pf_sql			nvarchar(max)<br />
	DECLARE @ps_sql			nvarchar(max)</p>
<p>	&#8211; Set Variable values<br />
	SET		@db_name		= &#8216;Tutorial&#8217;<br />
	SET		@fg_name		= &#8216;fg_date_monthly&#8217;<br />
	SET		@file_name		= &#8216;saledate_&#8217; + replace(convert(varchar(7),@nDate,121),&#8217;-',&#8221;)<br />
	SET		@ps_name		= &#8216;ps_date_monthly&#8217;<br />
	SET		@pf_name		= &#8216;pf_date_monthly&#8217;</p>
<p>	&#8211; DETERMINE DATABASE DATA PATHNAME<br />
	SELECT	@path = [filename] FROM master..sysdatabases WHERE NAME like @db_name<br />
	IF charindex(&#8216;\&#8217;, @path) &gt; 0<br />
	BEGIN<br />
		SELECT @path = substring(@path, 1, len(@path) &#8211; charindex(&#8216;\&#8217;, reverse(@path)))<br />
	END<br />
	SET		@path = @path</p>
<p>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-</p>
<p>	&#8211; ADD FILEGROUP<br />
	IF Not Exists (SELECT * FROM sys.filegroups WHERE NAME = @fg_name)<br />
	SET		@filegroup_sql =<br />
		&#8216;ALTER DATABASE &#8216; + QuoteName(@db_name,&#8221;) + &#8216; &#8216; + 	&#8216;ADD FILEGROUP &#8216; + QuoteName(@fg_name,&#8221;)<br />
	EXEC( @filegroup_sql )</p>
<p>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-</p>
<p>	&#8211; ADD FILE TO FILEGROUP<br />
	IF NOT EXISTS (SELECT * FROM sys.sysfiles WHERE NAME = @file_name)<br />
	SET		@file_sql =<br />
			&#8216;ALTER DATABASE&#8217; + &#8216; &#8216; + QuoteName(@db_name,&#8221;) + &#8216; &#8216; + &#8216;<br />
			 ADD FILE(	Name = &#8216; + QuoteName(@file_name,&#8221;&#8221;) + &#8216;,<br />
						FileName = &#8216; + QuoteName( @Path + &#8216;\&#8217; + @file_name + &#8216;.ndf&#8217;, &#8221;&#8221; ) + &#8216;,<br />
						SIZE = 3072 KB, MAXSIZE = UNLIMITED, FILEGROWTH = 1024 KB<br />
					  ) TO FILEGROUP &#8216; + @fg_name<br />
	EXEC( @file_sql )</p>
<p>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-</p>
<p>	&#8211; CREATE PARTITION FUNCTION<br />
	IF NOT EXISTS(SELECT * FROM sys.partition_functions WHERE NAME  = @pf_name)<br />
	SET		@pf_sql =<br />
			&#8216;CREATE PARTITION FUNCTION &#8216; +  QuoteName(@pf_name,&#8221;) + &#8216;(numeric(8,0)) &#8216; +<br />
			&#8216;AS RANGE RIGHT FOR VALUES (20110301,20110401)&#8217;<br />
	EXEC( @pf_sql )</p>
<p>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-</p>
<p>	&#8211; CREATE PARTITION SCHEME<br />
	IF NOT EXISTS(SELECT * FROM sys.partition_schemes WHERE NAME = @ps_name)<br />
	SET		@ps_sql =<br />
			&#8216;CREATE PARTITION SCHEME &#8216; + QuoteName(@ps_name,&#8221;) + &#8216; &#8216; +<br />
			&#8216;AS  PARTITION &#8216; +  QuoteName(@pf_name,&#8221;) + &#8216; ALL TO (&#8216; + @fg_name  + &#8216;)&#8217;<br />
	EXEC( @ps_sql )</p>
<p>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-</p>
<p>	DECLARE @table_sql	nvarchar(max)<br />
	&#8211; CREATE A TABLE tblsaledate TO USE PARTITION SCHEME<br />
	IF NOT EXISTS(SELECT * FROM sys.objects WHERE NAME = &#8216;tblsaledate&#8217; AND type = &#8216;U&#8217;)<br />
	SET		@table_sql =<br />
			&#8216;CREATE TABLE dbo.tblsaledate<br />
			(<br />
			 sale_id int not null identity(1,1),<br />
			 sale_date datetime not null,<br />
			 sale_date_id	numeric(8,0)<br />
			) ON &#8216; + QuoteName(@ps_name,&#8221;) + &#8216;(sale_date_id)&#8217;<br />
	EXEC( @table_sql )</p>
<p>	&#8211; POPULATE tblsaledate TABLE with data<br />
	DECLARE @startdate		datetime<br />
	DECLARE @enddate		datetime<br />
	DECLARE @thisdate		datetime<br />
	DECLARE @day_id			int<br />
	SET		@startdate		= &#8217;2011-03-01&#8242;<br />
	SET		@enddate		= &#8217;2011-06-01&#8242;<br />
	SET		@thisdate		= @startdate<br />
	SET		@day_id			= 1<br />
	WHILE	(@thisdate &lt;= @enddate)<br />
	BEGIN<br />
		INSERT INTO dbo.tblsaledate (sale_date, sale_date_id)<br />
		VALUES (@thisdate, CAST(convert(varchar(8),@thisdate,112) As int))<br />
		SET @day_id			= @day_id + 1<br />
		SET @thisdate		= DATEADD(day, 1, @thisdate)<br />
	END</p>
<p>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;</p>
<p>	&#8211; SET NEW FILE TO BE USED FOR NEXT SET OF DATA<br />
	DECLARE	@aps_sql		nvarchar(max)<br />
	SET		@aps_sql =<br />
			&#039;ALTER PARTITION SCHEME &#039; + QuoteName(@ps_name,&#039;&#039;) + &#039; NEXT USED &#039; + @fg_name<br />
	EXEC( @aps_sql )</p>
<p>	&#8211; ADD NEW RANGE FOR PARTITION<br />
	DECLARE @apf_sql		nvarchar(max)<br />
	SET		@apf_sql =<br />
			&#039;ALTER PARTITION FUNCTION &#039; + QuoteName(@pf_name,&#039;&#039;) + &#039;()<br />
			 SPLIT RANGE &#039; + &#039;(&#039; + CAST(@dateid as nvarchar) + &#039;)&#039;<br />
	EXEC( @apf_sql )</p>
<p>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;</p>
<p>	&#8211; IMPLEMENT ERROR HANDLING<br />
	SET		@ErrCount = @@ERROR<br />
	IF		(@ErrCount  0)<br />
	BEGIN<br />
		RETURN @cRtnFailed<br />
	END<br />
	SET NOCOUNT OFF;<br />
	RETURN	@cRtnSuccess</p>
<p>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211;<br />
/*</p>
<p>1. Data and Partition Views<br />
============================</p>
<p>	&#8212;-<br />
	SELECT * , $partition.pf_date_monthly(sale_date_id) FROM dbo.tblsaledate</p>
<p>	&#8212;-<br />
	SELECT * FROM sys.partitions WHERE object_id = object_id(&#8216;tblsaledate&#8217;)</p>
<p>2. Usage<br />
===================================<br />
	&#8212;-<br />
	Exec [dbo].[sp_dim_date_monthly_partition]</p>
<p>	&#8212;-<br />
	DROP PROCEDURE [dbo].[sp_dim_date_monthly_partition]<br />
	DROP PARTITION SCHEME [ps_date_monthly]<br />
	DROP PARTITION FUNCTION [pf_date_monthly]</p>
<p>*/<br />
&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;</p>
<p>END</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Kevin</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-130086</link>
		<dc:creator><![CDATA[Kevin]]></dc:creator>
		<pubDate>Wed, 20 Apr 2011 03:27:47 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-130086</guid>
		<description><![CDATA[HI PINALKUMAR DAVE,
When I created Partition for my database then : 

Partition function can only be created in Enterprise edition of SQL Server. Only Enterprise edition of SQL Server supports partitioning.

How can I solve with this situation? please explain to me 

I am very appreciated !

Kevin]]></description>
		<content:encoded><![CDATA[<p>HI PINALKUMAR DAVE,<br />
When I created Partition for my database then : </p>
<p>Partition function can only be created in Enterprise edition of SQL Server. Only Enterprise edition of SQL Server supports partitioning.</p>
<p>How can I solve with this situation? please explain to me </p>
<p>I am very appreciated !</p>
<p>Kevin</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: sharmi</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-122526</link>
		<dc:creator><![CDATA[sharmi]]></dc:creator>
		<pubDate>Wed, 09 Mar 2011 10:19:22 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-122526</guid>
		<description><![CDATA[how to create a partition function. And give the syntax pls]]></description>
		<content:encoded><![CDATA[<p>how to create a partition function. And give the syntax pls</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Will</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-121741</link>
		<dc:creator><![CDATA[Will]]></dc:creator>
		<pubDate>Thu, 03 Mar 2011 18:37:47 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-121741</guid>
		<description><![CDATA[Thank you for the excellent articles.  These have been a life saver several time.

I have partitioned my table by a DateTime field &quot;RecordTime&quot; and have partitioned it by MONTH

Two questions:

1.  Is there a query to retrieve the field name on which a table was partitioned ?

2.  Is there a query to retrieve the partition ranges?  In my example I am expecting to see the list of current partition ranges like Jan2011, Feb2011


Thanks.

Will]]></description>
		<content:encoded><![CDATA[<p>Thank you for the excellent articles.  These have been a life saver several time.</p>
<p>I have partitioned my table by a DateTime field &#8220;RecordTime&#8221; and have partitioned it by MONTH</p>
<p>Two questions:</p>
<p>1.  Is there a query to retrieve the field name on which a table was partitioned ?</p>
<p>2.  Is there a query to retrieve the partition ranges?  In my example I am expecting to see the list of current partition ranges like Jan2011, Feb2011</p>
<p>Thanks.</p>
<p>Will</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Mark Whitfield</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-119896</link>
		<dc:creator><![CDATA[Mark Whitfield]]></dc:creator>
		<pubDate>Mon, 21 Feb 2011 13:42:09 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-119896</guid>
		<description><![CDATA[Interesting thread...

I have a requirement to set-up an MS SQL Server 2005 (or 2008) database to hold significant amounts of transaction data. One table needs to hold 26 Gigabytes of data and the other, 3 Gigabytes. This is for data querying of the transaction data.

First question is &#039;Can it be done?&#039; in MS SQL Server.

Would I need to partition the tables over a number of hard drives to make this work?

What typical hardware configuration/approach would be recommended if the answer is YES?

Thanks in advance folks for any help, much appreciated...

Regards

Mark.]]></description>
		<content:encoded><![CDATA[<p>Interesting thread&#8230;</p>
<p>I have a requirement to set-up an MS SQL Server 2005 (or 2008) database to hold significant amounts of transaction data. One table needs to hold 26 Gigabytes of data and the other, 3 Gigabytes. This is for data querying of the transaction data.</p>
<p>First question is &#8216;Can it be done?&#8217; in MS SQL Server.</p>
<p>Would I need to partition the tables over a number of hard drives to make this work?</p>
<p>What typical hardware configuration/approach would be recommended if the answer is YES?</p>
<p>Thanks in advance folks for any help, much appreciated&#8230;</p>
<p>Regards</p>
<p>Mark.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: joel</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-119057</link>
		<dc:creator><![CDATA[joel]]></dc:creator>
		<pubDate>Fri, 18 Feb 2011 04:29:32 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-119057</guid>
		<description><![CDATA[hi, how can you resolve error like this. thanks.

ALTER TABLE SWITCH statement failed. The specified partition 13 of target table &#039;jfcdw_prd.dbo.FactSales_CK&#039; must be empty]]></description>
		<content:encoded><![CDATA[<p>hi, how can you resolve error like this. thanks.</p>
<p>ALTER TABLE SWITCH statement failed. The specified partition 13 of target table &#8216;jfcdw_prd.dbo.FactSales_CK&#8217; must be empty</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Zulfiqar</title>
		<link>http://blog.sqlauthority.com/2008/01/25/sql-server-2005-database-table-partitioning-tutorial-how-to-horizontal-partition-database-table/#comment-118045</link>
		<dc:creator><![CDATA[Zulfiqar]]></dc:creator>
		<pubDate>Fri, 11 Feb 2011 18:16:12 +0000</pubDate>
		<guid isPermaLink="false">http://sqlauthority.wordpress.com/?p=485#comment-118045</guid>
		<description><![CDATA[Hi Pinal
how to select from indivisual partition.
I mean i have one table that has four partitioning. 
Year1,Year2,Year3,Year4

how should I, select only one year1]]></description>
		<content:encoded><![CDATA[<p>Hi Pinal<br />
how to select from indivisual partition.<br />
I mean i have one table that has four partitioning.<br />
Year1,Year2,Year3,Year4</p>
<p>how should I, select only one year1</p>
]]></content:encoded>
	</item>
</channel>
</rss>

