跨多个表的SQL唯一约束

时间:2012-04-26 16:55:42

标签: sql-server database-design unique-constraint

我正在尝试跨多个表创建一个唯一约束。我在这里找到了类似的问题,但是他们并没有完全捕捉到我想要做的精神。

举个例子,我有三个表,t_Analog,t_Discrete,t_Message

CREATE TABLE t_Analog(
    [AppName] [nvarchar](20) NOT NULL,
    [ItemName] [nvarchar](32) NOT NULL,
    [Value] [float] NOT NULL,
    CONSTRAINT [uc_t_Analog] UNIQUE(AppName, ItemName)
)

CREATE TABLE t_Discrete(
    [AppName] [nvarchar](20) NOT NULL,
    [ItemName] [nvarchar](32) NOT NULL,
    [Value] [bit] NOT NULL,
    CONSTRAINT [uc_t_Discrete] UNIQUE(AppName, ItemName)
)

CREATE TABLE t_Message(
    [AppName] [nvarchar](20) NOT NULL,
    [ItemName] [nvarchar](32) NOT NULL,
    [Value] [nvarchar](256) NOT NULL,
    CONSTRAINT [uc_t_Message] UNIQUE(AppName, ItemName)
)

我的目标是让所有3个表中的AppName和ItemName都是唯一的。例如,应用程序X中的项目名称Y不能同时存在于模拟和离散表中。

请注意,这个例子是设计的,每个Type的实际数据都不同,大到足以组合表并添加一个非常难看的Type列。

如果您对此方法有任何建议,我很乐意听到它们!

----开始编辑2012-04-26 13:28 CST ----

谢谢大家的答案!

似乎可能有理由修改此数据库的架构,这很好。

将表组合到单个表中并不是一个可行的选项,因为每个类型的列数不超过30列(不幸的是,修改这些列不是一个选项)。这可能会导致每行中没有使用大部分列,这似乎是一个坏主意。

添加第4张表,如John Sikora和其他人提到的,可能是一个选项,但我想首先验证这一点。

将架构修改为:

CREATE TABLE t_AllItems(
    [id] [bigint] IDENTITY(1,1) NOT NULL,
    [itemType] [int] NOT NULL,
    [AppName] [nvarchar](20) NOT NULL,
    [ItemName] [nvarchar](32) NOT NULL,
    CONSTRAINT [pk_t_AllItems] PRIMARY KEY CLUSTERED ( [id] )
    CONSTRAINT [uc_t_AllItems] UNIQUE([id], [AppName], [ItemName])
) ON [PRIMARY]

CREATE TABLE t_Analog(
    [itemId] [bigint] NOT NULL,
    [Value] [float] NOT NULL,
    FOREIGN KEY (itemId) REFERENCES t_AllItems(id)
)

CREATE TABLE t_Discrete(
    [itemId] [bigint] NOT NULL,
    [Value] [bit] NOT NULL,
    FOREIGN KEY (itemId) REFERENCES t_AllItems(id)
)

CREATE TABLE t_Message(
    [itemId] [bigint] NOT NULL,
    [Value] [nvarchar](256) NOT NULL,
    FOREIGN KEY (itemId) REFERENCES t_AllItems(id)
)

我对这种方法只有一个问题。这是否会强制子表中的唯一性?

例如,是否存在“项目”,其中“id”9表格为t_Analog,其中“itemId”为9,“value”为9.3,同时t_Message的'itemId'9为'价值''foo“?

我可能不完全理解这种额外的表格方法,但我并不反对。

如果我错了,请纠正我。

6 个答案:

答案 0 :(得分:12)

专门为您想要独特的值添加第4个表,然后使用一对多关系将这些表中的这些键链接到其他表中。 例如,您将拥有一个唯一的表,其中包含ID,AppName和ItemName以构成其3列。然后将此表链接到其他人。

如何做到这一点就是一个很好的例子 Create a one to many relationship using SQL Server

编辑:这就是我要做的,但考虑到您的服务器需求,您可以更改所需内容:

CREATE TABLE AllItems(
    [id] [int] IDENTITY(1,1) NOT NULL,
    [itemType] [int] NOT NULL,
    [AppName] [nvarchar](20) NOT NULL,
    [ItemName] [nvarchar](32) NOT NULL,
    CONSTRAINT [pk_AllItems] PRIMARY KEY CLUSTERED ( [id] ASC )
) ON [PRIMARY]

CREATE TABLE Analog(
    [itemId] [int] NOT NULL,
    [Value] [float] NOT NULL
)

CREATE TABLE Discrete(
    [itemId] [int] NOT NULL,
    [Value] [bit] NOT NULL
)

CREATE TABLE Message(
    [itemId] [bigint] NOT NULL,
    [Value] [nvarchar](256) NOT NULL
)

ALTER TABLE [Analog] WITH CHECK 
    ADD CONSTRAINT [FK_Analog_AllItems] FOREIGN KEY([itemId])
REFERENCES [AllItems] ([id])
GO
ALTER TABLE [Analog] CHECK CONSTRAINT [FK_Analog_AllItems]
GO

ALTER TABLE [Discrete] WITH CHECK 
    ADD CONSTRAINT [FK_Discrete_AllItems] FOREIGN KEY([itemId])
REFERENCES [AllItems] ([id])
GO
ALTER TABLE [Discrete] CHECK CONSTRAINT [FK_Discrete_AllItems]
GO

ALTER TABLE [Message] WITH CHECK 
    ADD CONSTRAINT [FK_Message_AllItems] FOREIGN KEY([itemId])
REFERENCES [AllItems] ([id])
GO
ALTER TABLE [Message] CHECK CONSTRAINT [FK_Message_AllItems]
GO

从我可以告诉你的语法很好,我只是简单地改变它,因为我对它更熟悉,但要么应该工作。

答案 1 :(得分:9)

虽然您可能会或可能不想像其他答案那样改变您的架构,但indexed view可以应用您正在谈论的约束:

CREATE VIEW v_Analog_Discrete_Message_UK WITH SCHEMABINDING AS
SELECT a.AppName, a.ItemName
FROM dbo.t_Analog a, dbo.t_Discrete b, dbo.t_Message c, dbo.Tally t
WHERE (a.AppName = b.AppName and a.ItemName = b.ItemName)
    OR (a.AppName = c.AppName and a.ItemName = c.ItemName)
    OR (b.AppName = c.AppName and b.ItemName = c.ItemName)
    AND t.N <= 2
GO
CREATE UNIQUE CLUSTERED INDEX IX_AppName_ItemName_UK
    ON v_Analog_Discrete_Message_UK (AppName, ItemName)
GO

你需要一个"Tally" or numbers table或者必须动态生成一个,Celko风格:

-- Celko-style derived numbers table to 100k
select a.N + b.N * 10 + c.N * 100 + d.N * 1000 + e.N * 10000 + 1 as N
from (select 0 as N union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) a
      , (select 0 as N union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) b
      , (select 0 as N union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) c
      , (select 0 as N union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) d
      , (select 0 as N union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) e
order by N

答案 2 :(得分:1)

一种想法可能是将三个表组合在一起:

CREATE TABLE t_Generic(
[AppName] [nvarchar](20) NOT NULL,
[ItemName] [nvarchar](32) NOT NULL,
[Type] [nvarchar](32) NOT NULL,
[AnalogValue] [Float] NULL,
[DiscreteValue] [bit] NULL,
[MessageValue] [nvarchar](256) NULL,
CONSTRAINT [uc_t_Generic] UNIQUE(AppName, ItemName)
)

您的应用程序逻辑必须强制只填充一个值,并且您可以使用“类型”字段来跟踪该记录的类型。

答案 3 :(得分:1)

您还可以创建一个具有更多逻辑的约束并检查所有三个表。

请查看here,了解如何使用函数执行此操作。

答案 4 :(得分:0)

这会建议规范化/数据库设计问题,特别是你应该将appname存储在它自己的一个表中(作为唯一/密钥),然后是第二列,表示它链接到的ID,也许第3列表示类型。

EG:

AppName – PrimaryKey - unique
ID – Foreign Key of either Discrete, Analog or message
Type – SMALLINT representing Discrete, analog or message.

答案 5 :(得分:0)

我使用而不是插入和更新触发器来解决此问题,如下所示:

CREATE TRIGGER tI_Analog ON t_Analog
INSTEAD OF INSERT
AS 
BEGIN
    SET NOCOUNT ON ;

    IF EXISTS (SELECT 1 FROM inserted AS I INNER JOIN t_Analog AS T
                   ON T.AppName = I.AppName AND T.ItemName = I.ItemName
               UNION ALL
               SELECT 1 FROM inserted AS I INNER JOIN t_Discrete AS T
                   ON T.AppName = I.AppName AND T.ItemName = I.ItemName
               UNION ALL
               SELECT 1 FROM inserted AS I INNER JOIN t_Message AS T
                   ON T.AppName = I.AppName AND T.ItemName = I.ItemName
              )
    BEGIN
        RAISERROR('Duplicate key', 16, 10) ;
    END
    ELSE
    BEGIN
        INSERT INTO t_Analog ( AppName, ItemName, Value )
        SELECT AppName, ItemName, Value FROM inserted ;
    END
END
GO

CREATE TRIGGER tU_Analog ON t_Analog
INSTEAD OF UPDATE
AS 
BEGIN
    SET NOCOUNT ON ;

    IF EXISTS (SELECT TOP(1) 1
                 FROM (SELECT T.AppName, T.ItemName, COUNT(*) AS numRecs
                         FROM
                            (SELECT I.AppName, I.ItemName
                               FROM inserted AS I INNER JOIN t_Analog AS T
                                 ON T.AppName = I.AppName AND T.ItemName = I.ItemName
                             UNION ALL
                             SELECT I.AppName, I.ItemName
                               FROM inserted AS I INNER JOIN t_Discrete AS T
                                 ON T.AppName = I.AppName AND T.ItemName = I.ItemName
                             UNION ALL
                             SELECT I.AppName, I.ItemName
                               FROM inserted AS I INNER JOIN t_Message AS T
                                 ON T.AppName = I.AppName AND T.ItemName = I.ItemName
                            ) AS T
                          GROUP BY T.AppName, T.ItemName
                        ) AS T
                WHERE T.numRecs > 1
              )
    BEGIN
        RAISERROR('Duplicate key', 16, 10) ;
    END
    ELSE
    BEGIN
        UPDATE T
           SET AppName = I.AppName
             , ItemName = I.ItemName
             , Value = I.Value
          FROM inserted AS I INNER JOIN t_Message AS T
            ON T.AppName = I.AppName AND T.ItemName = I.ItemName
        ;
    END
END
GO

使用而不是触发器的一个警告是涉及到身份字段时。此触发器可防止INSERT INTO命令的OUTPUT子句和@@ IDENTITY变量正常工作。