Reputation: 148744
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.
Upvotes: 203
Views: 407732
Reputation: 867
If you don't want to include the backslash, but only the text after it, try this: ([^\\]+)$
or for unix: ([^\/]+)$
Upvotes: 12
Reputation: 16978
I used below regex to get that result also when its finished by a \
(\\[^\\]+)\\?$
Upvotes: 4
Reputation: 6105
One that worked for me was:
.+(\\.+)$
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
Upvotes: 142
Reputation: 2887
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.
Upvotes: 68
Reputation: 93086
Your negative lookahead solution would e.g. be this:
\\(?:.(?!\\))+$
See it here on Regexr
Upvotes: 161
Reputation: 143299
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.
Upvotes: 33