SQL Unique constraint across multiple tables

I am trying to create a unique constraint across multiple tables. I have found similar questions answered here but they don't quite capture the spirit of what I am trying to do.

As an example, I have three tables, 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)
)

My goal is to make AppName and ItemName unique across all 3 tables. For instance, an item name of Y in application X cannot exist in both analog and discrete tables.

Please note that this example is contrived, the actual data for each Type is different and large enough to make combining tables and adding a Type column pretty ugly.

If you have any suggestions on approaches to this, I would love to hear them!

---- BEGIN EDIT 2012-04-26 13:28 CST ----

Thank you all for your answers!

It seems there may be cause to modify the schema of this database, and that is fine.

Combining the tables into a single table is not really a viable option as there are on the order of 30 columns for each type that do not match (modifying these columns is, unfortunately, not an option). This could lead to large sections of columns not being used in each row, which seems like a bad idea.

Adding a 4th table, like John Sikora and others mention, may be an option but I would like to verify this first.

Modifying Schema to be:

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)
)

I only have one question regarding this approach. Does this enforce uniqueness across the sub tables?

For instance, could there not exist an 'Item' that has 'id' 9 with tables t_Analog having 'itemId' of 9 with 'value' of 9.3 and, at the same time, t_Message have 'itemId' 9 with 'Value' of "foo"?

I may not fully understand this extra table approach but I am not against it.

Please correct me if I am wrong on this.

Add a 4th table specifically for these values you want to be unique then link these keys from this table into the others using a one to many relationship. For example you will have the unique table with an ID, AppName and ItemName to make up its 3 columns. Then have this table link to the others.

For how to do this here is a good example Create a one to many relationship using SQL Server

EDIT: This is what I would do but considering your server needs you can change what is needed:

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

From what I can tell your syntax is fine, I simply changed it to this way simply because I am more familiar with it but either should work.

While you may or may not want to alter your schema like other answers say, an indexed view can apply the constraint that you're talking about:

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

You will need a "Tally" or numbers table or have to otherwise generate one on the fly, Celko-style:

-- 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

SQL Unique constraint across multiple tables

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