Oracle - 字符串組合排列

Question

我認爲我有一個複雜的需求。

這是一個使用Oracle 10.2的組合排列，我能夠使用笛卡爾連接來解決它，但我認爲它需要一些改進才能使它變得最簡單和更靈活。

主要行爲。

輸入字符串： '一兩個'

輸出： '一' '二' '二合一' '兩個一'

對於我的解決方案我已經限制字符串的數量爲5（注意輸出是一個接近階乘的數字）

SQL：

with My_Input_String as (select 1 as str_id, 'alpha beta omega gama' as str from dual) 

--------logic------- 

, String_Parse as (
        SELECT REGEXP_SUBSTR(str, '[^ ]+', 1, ROWNUM) str 
        FROM My_Input_String 
        where rownum < 6 -- string limitation -- 
        CONNECT BY level <= LENGTH(REGEXP_REPLACE(str, '([^ ])+|.', '\1')) 
       )  

--------CRAP select need refactoring------- 

select str from String_Parse 
union 
select REGEXP_REPLACE(trim(s1.str||' '||s2.str||' '||s3.str||' '||s4.str||' '||s5.str), '(){2,}', ' ') as str 
from 

(select str from String_Parse union select ' ' from dual) s1, 
(select str from String_Parse union select ' ' from dual) s2, 
(select str from String_Parse union select ' ' from dual) s3, 
(select str from String_Parse union select ' ' from dual) s4, 
(select str from String_Parse union select '  ' from dual) s5 
where 
-- 
s1.str <> s2.str and s1.str <> s3.str and s1.str <> s4.str and s1.str <> s5.str 
-- 
and s2.str <> s3.str and s2.str <> s4.str and s2.str <> s5.str 
-- 
and s3.str <> s4.str and s3.str <> s5.str 
-- 
and s4.str <> s5.str 

Answer 1

編輯：有一個通用的。最後真的很簡單（但花了我一段時間去那裏）

WITH words AS 
( SELECT REGEXP_SUBSTR('&txt', '\S+', 1, LEVEL) AS word 
     , LEVEL          AS num 
    FROM DUAL 
    CONNECT BY LEVEL <= LENGTH(REGEXP_REPLACE('&txt', '\S+\s*', 'X')) 
) 
SELECT SYS_CONNECT_BY_PATH(W.word, ' ') 
FROM words W 
CONNECT BY NOCYCLE PRIOR W.num != W.num 

Edit2：刪除多餘的maxnum東西。從以前的嘗試中遺留下來