使用regexp_substr在Oracle中按順序拆分字符串

Question

我在Oracle數據庫中有一個字符串，我的字符串是：'bbb; aaa; qqq; ccc'

我用正則表達式來分割我的字符串：

select distinct trim(regexp_substr('bbb;aaa;qqq;ccc','[^;]+', 1,level) ) as q 
from dual
connect by regexp_substr('bbb;aaa;qqq;ccc', '[^;]+', 1, level) is not null ;

我想按順序拆分它，我總是希望以下輸出：

bbb
aaa
qqq
ccc

因為subString的順序對我來說非常重要。 但是此查詢的結果不正確：

qqq
aaa
bbb
ccc

Answer 1

您不需要DISTINCT即可獲得結果； 此外，要獲得給定順序的結果，您只需要一個ORDER BY子句：

select trim(regexp_substr('bbb;aaa;qqq;ccc','[^;]+', 1,level) ) as q 
from dual
connect by regexp_substr('bbb;aaa;qqq;ccc', '[^;]+', 1, level) is not null
order by level

Answer 2

如果您確實需要DISTINCT ：

WITH your_data( value ) AS (
  SELECT 'bbb;aaa;qqq;ccc;aaa;eee' FROM DUAL
),
positions ( string, lvl, start_pos, end_pos ) AS (
  SELECT value, 1, 1, INSTR( value, ';', 1, 1 ) FROM your_data
UNION ALL
  SELECT string, lvl + 1, end_pos + 1, INSTR( string, ';', 1, lvl + 1 )
  FROM positions
  WHERE  end_pos > 0
),
substrings ( string, substring, lvl, start_pos ) AS (
  SELECT string,
         DECODE( end_pos, 0, SUBSTR( string, start_pos ), SUBSTR( string, start_pos, end_pos - start_pos ) ),
         lvl,
         start_pos
  FROM   positions
)
SELECT string,
       substring,
       lvl
FROM   substrings
WHERE  INSTR( ';' || string || ';', ';' || substring || ';' ) = start_pos;

輸出 ：

STRING                  SUBSTRING                      LVL
----------------------- ----------------------- ----------
bbb;aaa;qqq;ccc;aaa;eee bbb                              1
bbb;aaa;qqq;ccc;aaa;eee aaa                              2
bbb;aaa;qqq;ccc;aaa;eee qqq                              3
bbb;aaa;qqq;ccc;aaa;eee ccc                              4
bbb;aaa;qqq;ccc;aaa;eee eee                              6