如何添加外键关系?
我正在使用现有的 SQL 2005 数据库,该数据库未使用表之间的 FK 关系来实现。 我尝试添加与数据库图表的关系,但我的应用程序在尝试编辑或插入与新 FK 相关的任何数据时立即崩溃。
dbo.person [person_id | firstname | lastname | dateofbirth]
dbo.campaign [campaign_id | campaign_description]
dbo.disposition [disposition_id | disposition_description]
dbo.person_campaigns [person_campaign_id | person_id | campaign_id | disposition_id]
person_campaigns 表是将人员、营销活动和处置联系在一起的地方。 您能否提供适当的 SQL 语法来在这些实体之间添加正确的 FK 关系?
编辑
CREATE TABLE [dbo].[person_campaigns](
[person_campaigns_id] [int] IDENTITY(1,1) NOT NULL,
[person_id] [int] NOT NULL,
[d_campaign_id] [int] NOT NULL,
[d_physician_disposition_id] [int] NULL,
CONSTRAINT [PK_person_campaigns] PRIMARY KEY CLUSTERED
(
[person_campaigns_id] 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
CREATE TABLE [dbo].[d_campaign](
[d_campaign_id] [int] IDENTITY(1,1) NOT NULL,
[name] [varchar](50) NULL,
[year] [int] NULL,
[isactive] [bit] NOT NULL,
CONSTRAINT [PK_d_campaign] PRIMARY KEY CLUSTERED
(
[d_campaign_id] 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
GO
ALTER TABLE [dbo].[d_campaign] ADD CONSTRAINT [DF_d_campaign_isactive] DEFAULT ((1)) FOR [isactive]
GO
CREATE TABLE [dbo].[d_disposition](
[d_disposition_id] [int] IDENTITY(1,1) NOT NULL,
[name] [varchar](50) NULL,
[isactive] [bit] NOT NULL,
CONSTRAINT [PK_d_disposition] PRIMARY KEY CLUSTERED
(
[d_disposition_id] 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
GO
ALTER TABLE [dbo].[d_disposition] ADD CONSTRAINT [DF_d_disposition_isactive] DEFAULT ((1)) FOR [isactive]
GO
CREATE TABLE [dbo].[person](
[person_id] [int] IDENTITY(1,1) NOT NULL,
[firstname] [varchar](30) NULL,
[lastname] [varchar](30) NULL,
[dateofbirth] [datetime] NULL
CONSTRAINT [PK__person__0BC6C43E] PRIMARY KEY CLUSTERED
(
[person_id] 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
I'm working with an existing SQL 2005 database that was not implemented with FK relationships between tables. I tried to add the relationships with a database diagram and my application immediately blew up trying to edit or insert any data that is tied to the new FK.
dbo.person [person_id | firstname | lastname | dateofbirth]
dbo.campaign [campaign_id | campaign_description]
dbo.disposition [disposition_id | disposition_description]
dbo.person_campaigns [person_campaign_id | person_id | campaign_id | disposition_id]
The person_campaigns table is where a person, campaign, and disposition are tied together. Can you please provide the appropriate SQL syntax for adding the proper FK relationships between these entities?
EDIT
CREATE TABLE [dbo].[person_campaigns](
[person_campaigns_id] [int] IDENTITY(1,1) NOT NULL,
[person_id] [int] NOT NULL,
[d_campaign_id] [int] NOT NULL,
[d_physician_disposition_id] [int] NULL,
CONSTRAINT [PK_person_campaigns] PRIMARY KEY CLUSTERED
(
[person_campaigns_id] 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
CREATE TABLE [dbo].[d_campaign](
[d_campaign_id] [int] IDENTITY(1,1) NOT NULL,
[name] [varchar](50) NULL,
[year] [int] NULL,
[isactive] [bit] NOT NULL,
CONSTRAINT [PK_d_campaign] PRIMARY KEY CLUSTERED
(
[d_campaign_id] 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
GO
ALTER TABLE [dbo].[d_campaign] ADD CONSTRAINT [DF_d_campaign_isactive] DEFAULT ((1)) FOR [isactive]
GO
CREATE TABLE [dbo].[d_disposition](
[d_disposition_id] [int] IDENTITY(1,1) NOT NULL,
[name] [varchar](50) NULL,
[isactive] [bit] NOT NULL,
CONSTRAINT [PK_d_disposition] PRIMARY KEY CLUSTERED
(
[d_disposition_id] 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
GO
ALTER TABLE [dbo].[d_disposition] ADD CONSTRAINT [DF_d_disposition_isactive] DEFAULT ((1)) FOR [isactive]
GO
CREATE TABLE [dbo].[person](
[person_id] [int] IDENTITY(1,1) NOT NULL,
[firstname] [varchar](30) NULL,
[lastname] [varchar](30) NULL,
[dateofbirth] [datetime] NULL
CONSTRAINT [PK__person__0BC6C43E] PRIMARY KEY CLUSTERED
(
[person_id] 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
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由于我这台电脑上没有 SQL Server,而且我不记得语法,所以最简单的事情就是创建两个新的测试表,创建带有 ID 字段的 TableA,以及带有 TableA 的 FK 字段的 TableB .ID,然后编写 TableB 脚本以查看 ADD CONSTRAINT 语法。 使用 SQL Server Management Studio 通过数据库图表执行此操作。
但是,如果您能够在数据图中成功创建 FK,并且只能添加或更新记录,我相信还有其他问题。 编写 person_campaigns 表的脚本并发布代码。
Since I don't have SQL Server on this PC and I don't memorize the syntax, the easiest thing to do is to create two new test tables, create TableA with an ID field, TableB with a field that is a FK of TableA.ID, and then script out TableB to see the ADD CONSTRAINT syntax. Do this using SQL Server Management Studio via a database diagram.
However, if you were able to successfully create the FKs in a data diagram, and you only can't add or update records, I believe something else is wrong. Script out the person_campaigns table and post the code.
最简单的方法是通过数据库图表编辑器; 一次执行一个操作,并在每次连接完成后保存图表以影响表。 如果它“爆炸”,很可能是因为表包含不存在的外键值; 你必须先清理这些。
the easiest way to do it is through the database diagram editor; do them one at a time and save the diagram to affect the tables after each connection is made. If it "blows up" it is most likely because the tables contain foreign-key values that do not exist; you'll have to clean these up first.
如果您必须在创建表后添加它们,语法是
If you have to add them after the table is created the syntax is
假设我有两个应该有外键但没有的表。 首先要做的就是检查一下如果我设置了外键,是否会出现数据问题。
类似下面的代码将获取子表中与父表中没有匹配项的记录。
一旦您发现现有数据存在问题,您就需要创建一种方法来修复它。 修复方法将根据您拥有的与父表无关的数据以及表所代表的内容而有所不同。 假设您的父表包含汽车的 VIN 号作为 PK。 如果您的子表包含商店加工过的汽车,您可能希望通过将不存在的 VINS 添加到主表来解决问题,因为您不想丢失加工过的历史记录。 还有其他结构,您可能只想删除子表中不匹配的记录,因为它们没有意义。 在其他情况下,您可能希望将这些记录更新为某个默认值(可能是客户表中名为未知的客户)。 在其他情况下,您可能需要转到审核表或备份来查找已删除的 PK 的值,但未删除关联的子记录。 解决此问题的实际方法在很大程度上取决于数据的用途以及保留所有历史记录的重要性。 由于出于法律(和会计)原因,您永远不应该删除可能与金融交易相关的任何记录,因此您需要对这些记录非常小心。
修复所有数据后,运行代码来创建 FK 约束。
Suppose I had two tables that should have had a foreign key but did not. The first thing to do is check to see if there will be a data problem if I set a foreign key.
something like the below code would get you the records in the child table that do not have a match in the parent table.
Once you can see what is wrong with the existing data, then you need to create a way to fix it. The fix will vary depending on what data you have that has no relationship to the Parent table and what the tables represent. Suppose your parent table contained a VIN number for automobiles as the PK. If your child table contains the cars that were worked on by the shop, you would want to fix the issue by adding the nonexisting VINS to the primary table becasue you wouldn't want to lose the history of what was worked on. There are other structures where you might want to simply delete the records that don't match in child table because they are meaningless. In other circumstances you might want to update those records to some default value (perhaps a customer in the customer table called unknown). In still other circumstances, you might need to go to audit tables or backups to find the value of the PK that was deleted without the associated child records being deleted. The actual way to fix this problem is highly dependent on what the data is used for and how important it is to retain all historical records. Since you should never delete any record that might be related to a financial transaction for legal (and accounting) reasons, you need to be most careful with those.
After fixing all the data, then you run the code to create the FK constraint.