oracle中使用REGEXP_SUBSTR取值(最好用正则表达式)

2025-05-18 22:38:30
推荐回答(1个)
回答1:

select distinct regexp_substr('020101:1:13,020101:1:175,020301:1:107+020401:1:140-020501:1:107,020301:1:107*020401:1:140/020501:1:107',
'[^,|+|*|/|-]+', 1, level) as str
from dual
connect by level <= regexp_count('020101:1:13,020101:1:175,020301:1:107+020401:1:140-020501:1:107,020301:1:107*020401:1:140/020501:1:107',
'[,|+|*|/|-]') + 1
order by str;