Question

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

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

使用正确的GUID正确插入记录。

有关如何解决此问题的任何想法？我使用的是2.2.0.0版本

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

Answer 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都不知道如何执行此操作。鉴于guid是guid并且无关紧要谁生成它，我建议您在保存新记录之前使用UuidCreateSequential在客户端生成它。

Subsonic2.2和NEWSEQUENTIALID（）主键列

1 个答案: