RegEx:将所有内容匹配到最后一个空格而不包括它

Question

我想将字符串中的所有内容匹配到最后一个空格，但不包括它。例如，我想匹配我放入 中的字符。粗体 :

RENATA T. GROCHAL

So far I have ^(.+\s)(.+) However, it matches the last space and I don't want it to. RegEx should work also for other languages than English, as mine does.

EDIT: I didn't mention that the second capturing group should not contain a space – it should be GROCHAL not GROCHAL with a space before it.

EDIT 2: My new RegEx based on what the two answers have provided is: ^((.+)(?=\s))\s(.+) and the RegEx used to replace the matches is \3, \1. It does the expected result:

GROCHAL, RENATa T.

任何改进都是可取的。

Answer 1

^(.+)\s(.+)
带替换字符串:
\2, \1
更新:

另一个可以折叠 2 个捕获组之间的额外空间的版本:

^(.+?)\s+(\S+)$