Subsonic2.2 和 NEWSEQUENTIALID() 主键列答案

【问题标题】：Subsonic2.2 and NEWSEQUENTIALID() Primary Key ColumnSubsonic2.2 和 NEWSEQUENTIALID() 主键列
【发布时间】：2010-01-05 20:57:11
【问题描述】：

当我插入一条记录并尝试在插入后获取它的密钥时，Subsonic 返回 00000000-0000-0000-0000-000000000000。

product.Save();
GUID = product.ProdID;

使用正确的 GUID 正确插入记录。

关于如何解决这个问题的任何想法？我使用的是 2.2.0.0 版

这是我的表架构

GO

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
SET ANSI_PADDING ON
GO
CREATE TABLE [dbo].[ISA_810_ControlTracking](
    [ISAID] [uniqueidentifier] ROWGUIDCOL  NOT NULL CONSTRAINT     [DF_ISA_810_ControlTracking_ISAID]  DEFAULT (newsequentialid()),
    [ISA000_01_Authorization_Information_Qualifier] [varchar](2) NOT NULL,
    [ISA000_02_Authorization_Information] [varchar](10) NOT NULL,
    [ISA000_03_Security_Information_Qualifier] [varchar](2) NOT NULL,
    [ISA000_04_Security_Information] [varchar](10) NOT NULL,
    [ISA000_05_Interchange_Id_Qualifier] [varchar](2) NOT NULL,
    [ISA000_06_Interchange_Sender_Id] [varchar](15) NOT NULL,
    [ISA000_07_Interchange_Id_Qualifier] [varchar](2) NOT NULL,
    [ISA000_08_Interchange_Receiver_Id] [varchar](15) NOT NULL,
    [ISA000_09_Interchange_Date] [datetime] NOT NULL,
    [ISA000_10_Interchange_Time] [datetime] NOT NULL,
    [ISA000_11_Interchange_Control_Standards_Identifier] [varchar](1) NOT NULL,
    [ISA000_12_Interchange_Control_Version_Number] [varchar](5) NOT NULL,
    [ISA000_13_Interchange_Control_Number] [int] NOT NULL,
    [ISA000_14_Acknowledgment_Requested] [varchar](1) NOT NULL,
    [ISA000_15_Usage_Indicator] [varchar](1) NOT NULL,
    [ISA000_16_Component_Element_Separator] [varchar](1) NOT NULL,
    [IEA000_01_Number_Of_Included_Functional_Groups] [int] NOT NULL,
    [IEA000_02_Interchange_Control_Number] [int] NOT NULL,
 CONSTRAINT [PK_ISA_810_ControlTrackingIndex] PRIMARY KEY CLUSTERED 
(
    [ISAID] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, IGNORE_DUP_KEY = OFF,     ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY],
 CONSTRAINT [IX_ISA_810_ControlTracking] UNIQUE NONCLUSTERED 
(
    [ISA000_06_Interchange_Sender_Id] ASC,
    [ISA000_08_Interchange_Receiver_Id] ASC,
    [ISA000_13_Interchange_Control_Number] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, IGNORE_DUP_KEY = OFF,     ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY]
) ON [PRIMARY]

GO
SET ANSI_PADDING OFF

【问题讨论】：

标签： subsonic guid

【解决方案1】：

与 IDENTITY 类型不同，应用程序无法确定插入时生成的 GUID。虽然这可以通过使用 OUTPUT 子句在 T-SQL 中实现：INSERT ... OUTPUT inserted.$ROWGUIDCOL VALUES(...) 大多数 ORM 不知道如何执行此操作。鉴于 guid 是一个 guid，与谁生成它无关，我建议您在保存新记录之前在客户端生成它，使用 UuidCreateSequential。

【讨论】：