将来自同一 SQL 数据库中不同表的数据与表名组合在一起答案

【问题标题】：Combine Data from Different tables in same SQL Database along with the Table Name将来自同一 SQL 数据库中不同表的数据与表名组合在一起
【发布时间】：2019-10-28 09:29:27
【问题描述】：

我正在尝试从同一 SQL 数据库中的多个表中获取数据以及从中获取这些数据的表名。

    Select [timestamp]
      ,[User ID]
      ,[HCA Location Code]
      ,[HCA Lead]
      ,[PC Location Code]
      ,[Approval Administrator]
      ,[Time Sheet Admin_]
    FROM
    (SELECT * FROM [LIVE].[dbo].[LIVE Database1$User Setup]
    Union ALL
    SELECT * FROM [LIVE].[dbo].[LIVE_Database2$User Setup]
    Union ALL
    SELECT * FROM [LIVE].[dbo].[LIVE_Database3$User Setup]) data
    GROUP BY
      [timestamp]
      ,[User ID]
      ,[HCA Location Code]
      ,[HCA Lead]
      ,[PC Location Code]
      ,[Approval Administrator]
      ,[Time Sheet Admin_]
    HAVING count(*)!=2

   For value like below

【问题讨论】：

您的问题到底是什么？你什么都没问我们。如果您只是想添加表格表格，为什么不添加带有表格名称的列作为文字字符串？ 'LIVE Database1$User Setup' AS TableName.
请将示例数据添加到您的问题中，以使其成为可重现的问题。

标签： sql sql-server-2014 sql-query-store

【解决方案1】：

Select [User ID]
        ,[HCA Location Code]
        ,[HCA Lead]
        ,[PC Location Code]
    ,[Approval Administrator]
    ,[Time Sheet Admin_]
    , Table_Name
FROM
(
    Select [User ID]
            ,[HCA Location Code]
            ,[HCA Lead]
            ,[PC Location Code]
            ,[Approval Administrator]
            ,[Time Sheet Admin_]
            ,'Table1' AS Table_Name
    FROM [DB].[dbo].[UAT Table1$User Setup] as Tablename1
    Union ALL
    Select [User ID]
            ,[HCA Location Code]
            ,[HCA Lead]
            ,[PC Location Code]
            ,[Approval Administrator]
            ,[Time Sheet Admin_]
            ,'Table2' AS Table_Name
    FROM [DB].[dbo].[UAT Table2$User Setup]as Tablename1
    Union ALL
    Select [User ID]
            ,[HCA Location Code]
            ,[HCA Lead]
            ,[PC Location Code]
            ,[Approval Administrator]
            ,[Time Sheet Admin_]
            ,'Table3' AS Table_Name
    FROM [DB].[dbo].[Table3$User Setup]as Tablename1) data

GROUP BY
      [User ID]
      ,[HCA Location Code]
      ,[HCA Lead]
      ,[PC Location Code]
      ,[Approval Administrator]
      ,[Time Sheet Admin_]
      ,Table_Name
HAVING count(*)!=2

【讨论】：