Question

我遵循一个在线示例，说明如何使用这样的脚本创建文件表

  CREATE TABLE [dbo].[Users] AS FILETABLE ON [PRIMARY] FILESTREAM_ON [FGFS]
    GO

当我像这样运行查询时

SELECT * FROM sys.key_constraints WHERE parent_object_id=xxxxx
SELECT * FROM sys.index_columns WHERE object_id=yyyyy

第一个给我3个键（1PK + 2UQs）

我怎么知道哪个是是FILETABLE_PRIMARY_KEY_CONSTRAINT_NAME还是FILETABLE_STREAMID_UNIQUE_CONSTRAINT_NAME还是FILETABLE_FULLPATH_UNIQUE_CONSTRAINT_NAME？

Answer 1

就像任何表一样，您可以使用sp_help快速获取索引详细信息：

EXEC sp_help N'dbo.Users';

索引详细信息如下：

+-----------------------------+------------------------------------------------------+---------------------------+
|         index_name          |                  index_description                   |        index_keys         |
+-----------------------------+------------------------------------------------------+---------------------------+
| PK__Users__5A5B77D52FBBA93E | nonclustered, unique, primary key located on PRIMARY | path_locator              |
| UQ__Users__9DD95BAF3C1BB377 | nonclustered, unique, unique key located on PRIMARY  | stream_id                 |
| UQ__Users__A236CBB35BE6FD63 | nonclustered, unique, unique key located on PRIMARY  | parent_path_locator, name |
+-----------------------------+------------------------------------------------------+---------------------------+

以下查询将使用目录视图返回约束名称和列名称：

SELECT kc.name AS constraint_name, c.name AS column_name
FROM sys.filetables AS t
JOIN sys.key_constraints AS kc ON kc.parent_object_id = t.object_id
JOIN sys.index_columns AS ic ON ic.object_id = kc.parent_object_id AND ic.index_id = kc.unique_index_id
JOIN sys.columns AS c ON c.object_id = ic.object_id AND c.column_id = ic.column_id
WHERE t.object_id = OBJECT_ID(N'dbo.Users', 'U')
ORDER BY kc.name, ic.key_ordinal;

如何在创建的文件表中区分键

1 个答案: