Regex Last occurrence?

问题

I'm trying to catch the last part after the last backslash
I need the \Web_ERP_Assistant (with the \)

My idea was :

C:\Projects\Ensure_Solution\Assistance\App_WebReferences\Web_ERP_WebService\Web_ERP_Assistant


\\.+?(?!\\)      //  I know there is something with negative look -ahead `(?!\\)`

But I can't find it.

[Regexer Demo]

回答1:

Your negative lookahead solution would e.g. be this:

\\(?:.(?!\\))+$

See it here on Regexr

回答2:

One that worked for me was:

.+(\\.+)$

Try it online!

Explanation:

.+     - any character except newline
(      - create a group
 \\.+   - match a backslash, and any characters after it
)      - end group
$      - this all has to happen at the end of the string

回答3:

A negative look ahead is a correct answer, but it can be written more cleanly like:

(\\)(?!.*\\)

This looks for an occurrence of \ and then in a check that does not get matched, it looks for any number of characters followed by the character you don't want to see after it. Because it's negative, it only matches if it does not find a match.

回答4:

You can try anchoring it to the end of the string, something like \\[^\\]*$. Though I'm not sure if one absolutely has to use regexp for the task.

回答5:

What about this regex: \\[^\\]+$

回答6:

If you don't want to include the backslash, but only the text after it, try this: ([^\\]+)$ or for unix: ([^\/]+)$

回答7:

I used below regex to get that result also when its finished by a \

(\\[^\\]+)\\?$

[Regex Demo]

来源：https://stackoverflow.com/questions/8374742/regex-last-occurrence