REGEXP_SUBSTR所需的輸出

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“，” 。 我也希望最后一個字符串作為輸出 。 即三位來自“一，二，三”及五名來自“八項，十九歲，五”。 我該怎么做？？

Answer 1

如果只想獲取最后一個單詞而不檢查您的字符串是否符合特定模式：

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

---

對評論的回應

如何從第一個字符串中獲取字符串二和從第二個字符串中獲取十九？

regexp_substr函數的第四個參數是模式的出現。 因此，要獲取字符串中的第二個單詞，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

如果需要從字符串中提取每個單詞：

-- 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;

我不確定Oracle是否具有后顧之憂，但您也可以嘗試以下操作：

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