Oracle REGEXP_SUBSTR()函数(列转行\行转列)

1、开发过程经常遇到“列转行\行转列”问题,如下图

 2、列转行

2.1 现场数据库一般会有中文逗号,分号、、、、、等情况。转之前我们先进行替换
select 
  (SELECT REGEXP_REPLACE(m.ALTBE,'[;||;||,||、]', ',') AS NAMES FROM DUAL) 
from REG_AltAccItem m  WHERE  id = '141000013302990181'

2.2 列转行
SELECT REGEXP_SUBSTR('100000003,100000007,100000018,100000020,100000021,100000022,100000023,410000002,410000003,','[^,]+', 1, ROWNUM) AS code FROM  DUAL 
	CONNECT BY ROWNUM <= LENGTH(REGEXP_REPLACE('100000003,100000007,100000018,100000020,100000021,100000022,100000023,410000002,410000003,', '[^,]+', NULL)) + 1

3、行转列

参数解释
String:需要进行正则处理的字符串。
pattern:正则表达式。
position:起始位置(从字符串的第几个开始正则,默认为1,注:数据库中的字符串起始位置为1)。
occurrence:获取第几组通过正则表达式分割出来的组。
modifier:模式(‘i’不区分大小写,‘c’区分大小写。其中默认没‘c’)。

SELECT wm_concat(NAME) FROM (
	select NAME from CODE_REGALLLICCODE a INNER JOIN (
	SELECT REGEXP_SUBSTR('100000003,100000007,100000018,100000020,100000021,100000022,100000023,410000002,410000003,','[^,]+', 1, ROWNUM) AS code FROM  DUAL 
	CONNECT BY ROWNUM <= LENGTH(REGEXP_REPLACE('100000003,100000007,100000018,100000020,100000021,100000022,100000023,410000002,410000003,', '[^,]+', NULL)) + 1)  b ON a.CODE = b.code
)

 

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转载自blog.csdn.net/weixin_41542329/article/details/127300205