问题
Here is a funky matching that I need to accomplish.
A5.1.9.11.2
Needs to become:
A05.01.09.11.02
The amount of DOT sections will vary from none to many. And the letter "A" will always be there and always be 1 character.
I would like to use the regexp_replace() function in order to use this as a sorting mechanism. Thanks.
回答1:
Oracle SQL doesn't support lookaround assertions, which would be useful for this case:
s/([0-9](?<![0-9]))/0\1/g
You'll have to use at least two replacements:
REGEXP_REPLACE(REGEXP_REPLACE(col, '([0-9]+)', '0\1'), '0([0-9]{2})', '\1')
来源：https://stackoverflow.com/questions/12584261/oracle-sql-regexp-replace-matching