Oracle SQL Regexp_replace匹配 [英] Oracle SQL Regexp_replace matching

问题描述

这是我需要完成的时髦匹配.

Here is a funky matching that I need to accomplish.

A5.1.9.11.2

需要成为:

A05.01.09.11.02

DOT部分的数量从无到多.并且字母"A"将始终存在且始终为1个字符.

The amount of DOT sections will vary from none to many. And the letter "A" will always be there and always be 1 character.

我想使用regexp_replace()函数以便将其用作排序机制.谢谢.

I would like to use the regexp_replace() function in order to use this as a sorting mechanism. Thanks.

推荐答案

Oracle SQL不支持环视断言，在这种情况下这将很有用:

Oracle SQL doesn't support lookaround assertions, which would be useful for this case:

s/([0-9](?<![0-9]))/0\1/g

您必须至少使用两个替换项:

You'll have to use at least two replacements:

REGEXP_REPLACE(REGEXP_REPLACE(col, '([0-9]+)', '0\1'), '0([0-9]{2})', '\1')

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