REGEXP_SUBSTR所需輸出

Question

SELECT REGEXP_SUBSTR('one,two,three',',[^,]+') AS reg_result FROM DUAL; 

REG_RESULT  
,two   

SELECT REGEXP_SUBSTR('eight,nineteen,five',',[^,]+') AS reg_result FROM DUAL; 

REG_RESULT  
,nineteen 

我必須從結果中刪除「，」。另外我希望最後的字符串 作爲輸出。即來自'一，二，三'的三個 & 五個從'八，十九，五'。 我該怎麼做？

Answer 1

如果想只得到了最後一個字，不檢查，如果你的字符串是否符合特定的模式：

SQL> with t1 as(
    2 select 'one,two,three' as str from dual 
    3 ) 
    4 select regexp_substr(str, '([[:alpha:]]+)$') last_word 
    5 from t1 
    6 ; 

LAST_WORD 
--------- 
three 

---

迴應評論

如何從第一個得到字符串兩個&從第二個十九個？

regexp_substr函數的第四個參數是模式的出現。因此，要獲得字符串中的第二個字，我們可以使用regexp_substr(str, '[^,]+', 1, 2)

SQL> with t1 as(
    2  select 'one,two,three' as str from dual 
    3  ) 
    4 select regexp_substr(str, '[^,]+', 1, 2) as Second_Word 
    5  from t1; 

Second_Word 
--------- 
two 

如果有需要從你的字符串中提取的每一個字：

-- sample of data from your question 
SQL> with t1 as(
    2  select 'one,two,three' as str from dual union all 
    3  select 'eight,nineteen,five' from dual 
    4 ), -- occurrences of the pattern 
    5 occurrence as(
    6 select level as ps 
    7  from (select max(regexp_count(str, '[^,]+')) mx 
    8    from t1 
    9   ) s 
10 connect by level <= s.mx 
11 ) -- the query 
12 select str 
13  , regexp_substr(str, '[^,]+', 1, o.ps) word 
14  , o.ps as word_num 
15 from t1 t 
16  cross join occurrence o 
17 order by str 
18 ; 

STR     WORD   WORD_NUM 
------------------- ----------- ---------- 
eight,nineteen,five eight    1 
eight,nineteen,five nineteen    2 
eight,nineteen,five five     3 
one,two,three  three    3 
one,two,three  one     1 
one,two,three  two     2 

6 rows selected 

Answer 2

SELECT REGEXP_SUBSTR(REGEXP_SUBSTR('one,two,three',',[^,]+$'),'[^,]+') AS reg_result FROM DUAL; 

我不能肯定，如果Oracle有lookbehinds，但你可以試試這個還有：

SELECT REGEXP_SUBSTR('one,two,three','(?<=,)[^,]+$') AS reg_result FROM DUAL;