您可以在条件准则中使用数据库名称和UNION运算子，从右边的表格中选取。

SELECT COLUMN_1, COLUMN_2, ..., COLUMN_N
    FROM
    (
     SELECT COLUMN_1, COLUMN_2, ..., COLUMN_N
     FROM TableA
     WHERE ora_database_name = 'DatabaseA'
     UNION ALL
     SELECT COLUMN_1, COLUMN_2, ..., COLUMN_N
     FROM TableB
     WHERE ora_database_name = 'DatabaseB'
     UNION ALL
     SELECT 'ERROR', null, ..., null
     FROM DUAL
     WHERE ora_database_name NOT IN ('DatabaseA', 'DatabaseB')
     )
     WHERE
     (my where criteria)

如果是从两个表中选择一个表的情况，则类似下面的内容可能会有所帮助：

WITH
    tab_a AS
        (   Select COL_1, COL_2, COL_3 From TABLE_A  ),   -- It is possible to do the filter here or in the main SQL or both
    tab_b AS
        (   Select COL_1, COL_2, COL_3 From TABLE_B  )    -- It is possible to do the filter here or in the main SQL or both

Select DISTINCT
    CASE WHEN 1 = 1 THEN a.COL_1 ELSE b.COL_1 END "COL_1",
    CASE WHEN 1 = 1 THEN a.COL_2 ELSE b.COL_2 END "COL_2",
    CASE WHEN 1 = 1 THEN a.COL_3 ELSE b.COL_3 END "COL_3"
From
    tab_a a
Inner Join
    tab_b b ON(1 = 1)
Where
    CASE WHEN 1 = 1 THEN b.COL_1 ELSE a.COL_2 END = 'XXX'   -- filter could be set on the same columns or on different ones of the same type

您可以将where子句放在cte定义（WITH子句）或主SQL中，或者同时放在这两个语句中。

• CASE表达式中的WHEN条件应根据数据上下文放置，这是从一个或另一个表中选择/筛选数据的位置。
• INNER JOIN ON条件可以保持原样（1=1），因为它会创建将由DISTINCT关键字解决的交叉链接。

如果两个表都不能从两个数据库访问，那么你应该有一个单独的脚本，每个数据库一个脚本。但是，你可以检查哪个是活动的数据库和/或是否有一个特定的表。

-- to get the db name ...
Select NAME "DB_NAME" From V$database

-- ... to check if the table exist ...
Select TABLE_NAME from ALL_TABLES where TABLE_NAME = 'TABLE_A'

-- ... or some additional info besides the table name
Select
    u.USER_ID, u.USERNAME,
    t.TABLE_NAME, t.OWNER "TABLE_OWNER"
From ALL_USERS u
Inner Join ALL_TABLES t ON(t.OWNER = u.USERNAME)
Where TABLE_NAME = 'TABLE_A'