这样做：

WITH innermost AS (SELECT 2)
SELECT * FROM innermost
UNION SELECT 3;

我得到这个：

?column? 
----------
        2
        3

这样做：

WITH outmost AS (
        (WITH innermost AS (SELECT 2)
         SELECT * FROM innermost)
)                                
SELECT * FROM outmost;

结果：

?column? 
----------
        2

这也可以：

WITH outmost AS (
  SELECT 1
  UNION (WITH innermost AS (SELECT 2)
         SELECT * FROM innermost)
)
SELECT * FROM outmost;

我得到这个：

?column? 
----------
        1
        2

但这不行：

WITH outmost AS (
  SELECT 1
  UNION (WITH innermost as (SELECT 2)
         SELECT * FROM innermost
         UNION SELECT 3)
)
SELECT * FROM outmost;

结果：

ERROR:  relation "innermost" does not exist
LINE 4:          SELECT * FROM innermost

对于我的思维方式，最后一个应该成功，还有一个应该失败。我看不到图案。有没有一些一般规则，可以使我预测嵌套CTE和UNION的组合将会或不会起作用？