先准备测试表
CREATE TABLE [dbo].[Employee] ( EmployeeNo INT IDENTITY(1,1) PRIMARY KEY, EmployeeName NVARCHAR(50) NULL );
插入一些数据
INSERT Employee DEFAULT VALUES; GO 10000 UPDATE Employee set EmployeeName = ‘Employee ‘ + CONVERT(VARCHAR, EmployeeNo)
创建文件组
创建文件(为了性能,可以设定文件Path分布在不同的磁盘分区上)
创建分区准备
选择分区列
创建分区函数
创建分区框架
定义边界值,分区,因为有5个边界值,所以需6个分区
产生的脚本文件(换句话说上面的步骤等同于下面的语句)
USE [TestingDB] GO BEGIN TRANSACTION CREATE PARTITION FUNCTION [EmpFunction](int) AS RANGE RIGHT FOR VALUES (N‘2000‘, N‘4000‘, N‘6000‘, N‘8000‘, N‘10000‘) CREATE PARTITION SCHEME [FunScheme] AS PARTITION [EmpFunction] TO ([FileGroup1], [FileGroup2], [FileGroup3], [FileGroup4], [FileGroup5], [PRIMARY]) ALTER TABLE [dbo].[Employee] DROP CONSTRAINT [PK__Employee__7AD0F1B633D4B598] ALTER TABLE [dbo].[Employee] ADD PRIMARY KEY CLUSTERED ( [EmployeeNo] ASC )WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, SORT_IN_TEMPDB = OFF, IGNORE_DUP_KEY = OFF, ONLINE = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [FunScheme]([EmployeeNo]) COMMIT TRANSACTION
执行上面的SQL语句
其中语句ON [FunScheme]([EmployeeNo])是关键,表明了表Employee依赖分区框架FunScheme来进行分区,分区的列为EmployeeNo
而分区框架有依赖于分区函数,即分区表依赖分区框架,分区框架又依赖于分区函数
查看表分区结果
时间: 2024-11-10 11:23:13