查找视图中使用的别名的真实列名？

Question

假设我有一个视图,其中一些列名是别名,例如本例中的"surName":

CREATE VIEW myView AS
    SELECT  
            firstName,
            middleName,
            you.lastName surName
    FROM 
            myTable me
            LEFT OUTER JOIN yourTable you
            ON me.code = you.code
GO

我可以使用INFORMATION_SCHEMA视图检索有关视图的一些信息.
例如,查询

SELECT column_name AS ALIAS, data_type AS TYPE
FROM information_schema.columns 
WHERE table_name = 'myView'

收益率:

 ----------------
|ALIAS     |TYPE |
 ----------------
|firstName |nchar|
|middleName|nchar|
|surName   |nchar|
 ----------------

但是,我想知道实际的列名.理想的情况是:

 ---------------------------
|ALIAS     |TYPE |REALNAME  |
 ---------------------------
|firstName |nchar|firstName |
|middleName|nchar|middleName|
|surName   |nchar|lastName  |
 ---------------------------

如何根据别名确定实际列名称是什么？ 必须有一些方法来使用sys表和/或INFORMATION_SCHEMA视图来检索此信息.

---

编辑: 我可以接受这种令人厌恶的东西,这与Arion的答案相似:

SELECT
    c.name AS ALIAS,
    ISNULL(type_name(c.system_type_id), t.name) AS DATA_TYPE,
    tablecols.name AS REALNAME
FROM 
    sys.views v
    JOIN sys.columns c ON c.object_id = v.object_id
    LEFT JOIN sys.types t ON c.user_type_id = t.user_type_id
    JOIN sys.sql_dependencies d ON d.object_id = v.object_id 
        AND c.column_id = d.referenced_minor_id
    JOIN sys.columns tablecols ON d.referenced_major_id = tablecols.object_id 
        AND tablecols.column_id = d.referenced_minor_id 
        AND tablecols.column_id = c.column_id
WHERE v.name ='myView'

这会产生:

 ---------------------------
|ALIAS     |TYPE |REALNAME  |
 ---------------------------
|firstName |nchar|firstName |
|middleName|nchar|middleName|
|surName   |nchar|code      |
|surName   |nchar|lastName  |
 ---------------------------

但是第三条记录是错误的 - 这种情况发生在使用"JOIN"子句创建的任何视图中,因为有两列具有相同的"column_id",但在不同的表中.

Answer 1

鉴于这种观点:

CREATE VIEW viewTest
AS
SELECT
    books.id,
    books.author,
    Books.title AS Name
FROM
    Books

我可以看到你可以获得使用的列和执行此操作所使用的表:

SELECT * 
FROM INFORMATION_SCHEMA.VIEW_COLUMN_USAGE AS UsedColumns 
WHERE UsedColumns.VIEW_NAME='viewTest'

SELECT * 
FROM INFORMATION_SCHEMA.VIEW_TABLE_USAGE AS UsedTables 
WHERE UsedTables.VIEW_NAME='viewTest'

这适用于sql server 2005+.见这里的参考

编辑

给出相同的观点.试试这个查询:

SELECT
    c.name AS columnName,
    columnTypes.name as dataType,
    aliases.name as alias
FROM 
sys.views v 
JOIN sys.sql_dependencies d 
    ON d.object_id = v.object_id
JOIN .sys.objects t 
    ON t.object_id = d.referenced_major_id
JOIN sys.columns c 
    ON c.object_id = d.referenced_major_id 
JOIN sys.types AS columnTypes 
    ON c.user_type_id=columnTypes.user_type_id
    AND c.column_id = d.referenced_minor_id
JOIN sys.columns AS aliases
    on c.column_id=aliases.column_id
    AND aliases.object_id = object_id('viewTest')
WHERE
    v.name = 'viewTest';

它为我返回:

columnName  dataType  alias

id          int       id
author      varchar   author
title       varchar   Name

这也在sql 2005+中测试过

Answer 2

我花了几个小时试图找到这个问题的答案，并反复遇到不起作用的解决方案和似乎最终放弃的海报，我最终在这里偶然发现了一个似乎有效的答案：

https://social.msdn.microsoft.com/Forums/windowsserver/en-US/afa2ed2b-62de-4a5e-ae70-942e75f887a1/find-out-original-columns-name-when-used-in-a-view- with-alias?forum=transactsql

我相信，下面的 SQL 返回的正是您正在寻找的内容，它确实满足了我的需要，并且看起来也表现良好。

SELECT  name
    , source_database
    , source_schema
    , source_table
    , source_column
    , system_type_name
    , is_identity_column
FROM    sys.dm_exec_describe_first_result_set (N'SELECT * from ViewName', null, 1) 

有关该函数的文档sys.dm_exec_describe_first_result_set可以在此处找到，它在 SQL Server 2012 及更高版本中可用：

https://learn.microsoft.com/en-us/sql/relational-databases/system-dynamic-management-views/sys-dm-exec-describe-first-result-set-transact-sql

完全归功于链接上的海报，我自己没有解决这个问题，但我想在这里发布这个，以防它对搜索此信息的其他人有用，因为我发现这个线程比我链接到的线程容易得多。

Answer 3

我想你不能.

选择查询会隐藏它所执行的实际数据源.因为你可以查询任何东西,即查看,表,甚至链接远程服务器.