SQL Server Partitioning - Unique Index Error

问题

I have a table that is partitioned by TRANSACTION_DATE_TIME.

Table has a column: ID.

I want to create a unique index for ID on partition scheme as:

CREATE UNIQUE NONCLUSTERED INDEX [IX_ID_ON_PS_DATETIME] ON [CRD].[TRANSACTION] 
(
    [ID] ASC
) ON [PS_DATETIME_WEEKLY]([TRANSACTION_DATE_TIME])

but SQL says "Partition column for a unique index must be a subset of index key".

I really don't need TRANSACTION_DATE_TIME column in this index.

How can I create the index without using TRANSACTION_DATE_TIME column?

回答1:

Or you create NON-partitioned index, or you HAVE to include the partitioning key into partitioned index like this:

Partitioned index

CREATE UNIQUE NONCLUSTERED INDEX [IX_ID_ON_PS_DATETIME] ON [CRD].[TRANSACTION] 
(
    [ID] ASC,
    TRANSACTION_DATE_TIME
) ON [PS_DATETIME_WEEKLY]([TRANSACTION_DATE_TIME])

OR

Non-partitioned index

CREATE UNIQUE NONCLUSTERED INDEX [IX_ID_ON_PS_DATETIME] ON [CRD].[TRANSACTION] 
(
    [ID] ASC
) ON PRIMARY

回答2:

From the TechNet Microsoft page

When partitioning a unique index (clustered or nonclustered), the partitioning column must be chosen from among those used in the unique index key. If it is not possible for the partitioning column to be included in the unique key, you must use a DML trigger instead to enforce uniqueness.

So to implement a workaround you can check this link out... one workaround from the blog and another one in the comments

Dealing with Unique Columns when Using Table Partitioning

来源：https://stackoverflow.com/questions/8710173/sql-server-partitioning-unique-index-error