在 Oracle 中使用 regexp_substr 按顺序拆分字符串

Posted 2023-03-27

【中文标题】在 Oracle 中使用 regexp_substr 按顺序拆分字符串

【英文标题】：Split string in Oracle with regexp_substr in order

【发布时间】：2017-05-22 07:05:05

【问题描述】：

我在Oracle数据库中有一个字符串，我的字符串是：'bbb;aaa;qqq;ccc'

我使用正则表达式来拆分我的字符串：

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 ;

我想按顺序拆分它，我希望总是下面的输出：

bbb
aaa
qqq
ccc

因为子字符串的顺序对我来说非常重要。但是这个查询的结果不按顺序：

qqq
aaa
bbb
ccc

【参考方案1】：

您不需要DISTINCT 即可获得结果；此外，要按给定顺序获得结果，您只需要一个 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

【参考方案2】：

如果你确实需要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;

输出：

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