Partición de tablas y tiempo de compilación de un plan de consulta en SQL Server

Para los futuros estudiantes del curso "Desarrollador de MS SQL Server", preparamos una traducción de un artículo útil.



También invitamos a todos a un seminario
web abierto sobre el tema "Polybase: la vida antes y después" . En el seminario web, veremos cómo era posible interactuar con otras bases de datos antes de Polybase y cómo funciona ahora.






: « SQL Server , ?»





, , , . . , : «».





, - Stack Overflow , :





USE StackOverflow;
GO

/* Create date partition function by day since Stack Overflow's origin,
modified from Microsoft Books Online: 
https://docs.microsoft.com/en-us/sql/t-sql/statements/create-partition-function-transact-sql?view=sql-server-ver15#BKMK_examples
 
DROP PARTITION SCHEME [DatePartitionScheme];
DROP PARTITION FUNCTION [DatePartitionFunction];
*/
DECLARE @DatePartitionFunction nvarchar(max) = 
    N'CREATE PARTITION FUNCTION DatePartitionFunction (datetime) 
    AS RANGE RIGHT FOR VALUES (';  
DECLARE @i datetime = '2008-06-01';
WHILE @i <= GETDATE()
BEGIN  
SET @DatePartitionFunction += '''' + CAST(@i as nvarchar(20)) + '''' + N', ';  
SET @i = DATEADD(DAY, 1, @i);  
END  
SET @DatePartitionFunction += '''' + CAST(@i as nvarchar(20))+ '''' + N');';  
EXEC sp_executesql @DatePartitionFunction;  
GO  
 
/* Create matching partition scheme, but put everything in Primary: */
CREATE PARTITION SCHEME DatePartitionScheme  
AS PARTITION DatePartitionFunction  
ALL TO ( [PRIMARY] ); 
GO
      
      



Users, CreationDate:





DROP TABLE IF EXISTS dbo.Users_partitioned;
GO
CREATE TABLE [dbo].[Users_partitioned](
	[Id] [int] NOT NULL,
	[AboutMe] [nvarchar](max) NULL,
	[Age] [int] NULL,
	[CreationDate] [datetime] NOT NULL,
	[DisplayName] [nvarchar](40) NOT NULL,
	[DownVotes] [int] NOT NULL,
	[EmailHash] [nvarchar](40) NULL,
	[LastAccessDate] [datetime] NOT NULL,
	[Location] [nvarchar](100) NULL,
	[Reputation] [int] NOT NULL,
	[UpVotes] [int] NOT NULL,
	[Views] [int] NOT NULL,
	[WebsiteUrl] [nvarchar](200) NULL,
	[AccountId] [int] NULL
) ON [PRIMARY];
GO
 
CREATE CLUSTERED INDEX CreationDate_Id ON 
	dbo.Users_partitioned (Id)
	ON DatePartitionScheme(CreationDate);
GO
 
INSERT INTO dbo.Users_partitioned (Id, AboutMe, Age,
	CreationDate, DisplayName, DownVotes, EmailHash,
	LastAccessDate, Location, Reputation, UpVotes,
	Views, WebsiteUrl, AccountId)
SELECT Id, AboutMe, Age,
	CreationDate, DisplayName, DownVotes, EmailHash,
	LastAccessDate, Location, Reputation, UpVotes,
	Views, WebsiteUrl, AccountId
	FROM dbo.Users;
GO
Let’s c
      
      



Users Users_partitioned. , Users_partitioned , , , :





CREATE INDEX DisplayName ON dbo.Users(DisplayName);
CREATE INDEX DisplayName ON dbo.Users_partitioned(DisplayName);
      
      



, :





SET STATISTICS TIME, IO ON;
SELECT * FROM dbo.Users WHERE DisplayName = N'Brent Ozar';
GO
SELECT * FROM dbo.Users_partitioned WHERE DisplayName = N'Brent Ozar';
GO
      
      



, , , , 0% , — 100%:





, 0,001, — 15. , (compile time), (execution time) (logical reads) . , — ( ):





27 . , : « 27 ?» — ! , 250 . , . - .





, , ETL- . , 250 .





, ?

Users_partitioned . , ON PRIMARY , .





CREATE INDEX DisplayName ON dbo.Users(DisplayName);
CREATE INDEX DisplayName ON dbo.Users_partitioned(DisplayName) ON [PRIMARY];
      
      



:





SET STATISTICS TIME, IO ON;
SELECT * FROM dbo.Users WHERE DisplayName = N'Brent Ozar';
GO
SELECT * FROM dbo.Users_partitioned WHERE DisplayName = N'Brent Ozar';
GO
      
      



:





- . , — :





, -

, , . (, rowstore- 100 ), , , . , , .





— , : «, , !» , : « , ».






"MS SQL Server Developer".





«Polybase: ».








All Articles