oracle 多个with as

今天去面试，遇到一个问题。

题目是这样的：

表1：

SQL> select * from tb_data;
 
        ID      CODE1      CODE2
---------- ---------- ----------
         1          1          2
         2          2          1
         3          2          3
         4          3          1
         5          4          2
         6          5          3
 
已选择6行。
 
SQL>

表2：

SQL> select * from tb_code;
 
      CODE DESCRIPTION
---------- --------------------
         1 a
         2 b
         3 c
         4 d
         5 e
 
SQL>

要求显示成：

        ID DESCRIPTION          DESCRIPTION
---------- -------------------- --------------------
         1 a                    b
         2 b                    a
         3 b                    c
         4 c                    a
         5 d                    b
         6 e                    c

也就是把表1里的code翻译成表2里的description。

因为面试官就坐在旁边，以前没写过，也没怎么动脑筋去写。

现在回来后，就想了一下。可以用多个with as来解决。

WITH
T3 AS
(
SELECT T1.ID, T1.CODE1, T2.DESCRIPTION
FROM TB_DATA T1, TB_CODE T2
WHERE T1.CODE1 = T2.CODE
),
T4 AS
(
SELECT T1.ID, T1.CODE2, T2.DESCRIPTION
FROM TB_DATA T1, TB_CODE T2
WHERE T1.CODE2 = T2.CODE
)
SELECT T3.ID, T3.DESCRIPTION, T4.DESCRIPTION
FROM T3, T4
WHERE T3.ID = T4.ID
ORDER BY ID;

执行后：

SQL> WITH
  2  T3 AS
  3  (
  4  SELECT T1.ID, T1.CODE1, T2.DESCRIPTION
  5  FROM TB_DATA T1, TB_CODE T2
  6  WHERE T1.CODE1 = T2.CODE
  7  ),
  8  T4 AS
  9  (
 10  SELECT T1.ID, T1.CODE2, T2.DESCRIPTION
 11  FROM TB_DATA T1, TB_CODE T2
 12  WHERE T1.CODE2 = T2.CODE
 13  )
 14  SELECT T3.ID, T3.DESCRIPTION, T4.DESCRIPTION
 15  FROM T3, T4
 16  WHERE T3.ID = T4.ID
 17  ORDER BY ID;
 
        ID DESCRIPTION          DESCRIPTION
---------- -------------------- --------------------
         1 a                    b
         2 b                    a
         3 b                    c
         4 c                    a
         5 d                    b
         6 e                    c
 
已选择6行。
 
SQL>