正则表达式捕获部分文本

Question

Could you help me, please! I have a string like this:

C:\\Temp dir\\Some one else\\File name.ext

I need to capture File name.ext But I can't understand how do it by regex.

When I trying to create regexp I can't understand how to capture only last symbol \\ and a text between them and .

Answer 1

Hmm.. the following RegEx should do it:

\\[^\\]*\.

Depending on your language and environment, further escaping may be required for the backslashes and/or other characters.

Answer 2

Try something like:

[^\\]*\.\w+$

Use $ to anchor to end.

Breakdown explanation of the regex:

[^\\]*   Match the file name. (Any characters other than a \)
\.       Match the '.'.
\w+      Match the extension.
$        Anchor to end.

Answer 3

In response to comments on other questions, here is an alternative.

If you want to capture text between backslash 6 and 7 you can do the following:

(?:[^\\]*\\){6}([^\\]*)

Explanation:

(?: ){6}   A non-capturing group repeated 6 times
[^\\]*     Any number of characters other than backslash
\\.        Followed by backslash
([^\\]*)   capture all non backslash characters that follow

You can see this at work (with better explanation) at http://regex101.com/r/fP2cL7

If you want a match between backslash 4 and 7, it would be

    (?:[^\\]*\\){4}((?:[^\\]*\\){2}[^\\]*)

As above but "don't do anything with the first four backslashes", then capture "no backslash followed by backslash twice, followed by no backslash one more time". Which captures everything from backslash 4 to backslash 7 (without including the slashes at the end)

Answer 4

Using Powershell

(gci $filepath).Name

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Using Regex

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

This will match everything after the last slash \\