Regex - get the fileName?

Question

I have this urls:

C:\Projects\Ensure_Solution\GD_EServices_Web\App_WebReferences\GD_Eservices_Web_Service\GD_Eservices_Web_Service.wsdl
C:\Projects\Ensure_Solution\GD_EServices_Web\App_WebReferences\GD_Eservices_Web_Service\GD_Eservices_Web_Service.wsdl

I want to get the wsdl file name ( no leading slash)

I have succeeded with 2 solutions :

\\[^\\]+$

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

But this returns the leading slash : http://regexr.com?32lvi

enter image description here

how can I enhance my regex return only the file ?

`Please include a tag specifying the programming language or tool you are using, together with this tag,` says the `regex` tag description. — Lev Levitsky, Nov 01 '12 at 11:06

score 1 · Answer 1 · answered Nov 01 '12 at 11:09

1

This should work [^\\]+$ But for your case I'd prefer smth like string.split('/').pop() (javascript) or array_pop(split('/', string)) (for php, I don't know language you are using) not regexp.

answered Nov 01 '12 at 11:09

Vyacheslav Voronchuk

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I know this solution `.split('/').pop();` but the question tagged as regex.... :-) – Royi Namir Nov 01 '12 at 11:10

score 1 · Answer 2 · edited Nov 01 '12 at 11:11

1

Try with the negative look-ahead (?!\\)(.(?!\\))+$

edited Nov 01 '12 at 11:11

Christoph

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answered Nov 01 '12 at 11:10

0x41ndrea

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js supports netgative look ahead ? or is it negative look behind i dont remember – Royi Namir Nov 01 '12 at 11:11
@RoyiNamir lookbehinds of any case are not supported, lookaheads are. – Christoph Nov 01 '12 at 11:16

score 1 · Accepted Answer · answered Nov 01 '12 at 11:13

You just need to exclude the the leading slash in the regex.

var path = 'C:\\Projects\\Ensure_Solution\\GD_EServices_Web\\App_WebReferences\\GD_Eservices_Web_Service\\GD_Eservices_Web_Service.wsdl';
console.log(path.match(/[^\\]+$/));

And you could get it without regex, use split, and get the last element with pop:

console.log(path.split('\\').pop());

score 0 · Answer 4 · edited Nov 01 '12 at 11:11

0

Try this

\\[^\\]+$

Note:that means try with only one leading backslash

edited Nov 01 '12 at 11:11

Christoph

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answered Nov 01 '12 at 11:08

Flow

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score 0 · Answer 5 · edited May 23 '17 at 12:12

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This should do it:

([\w\d_-]*)\.?[^\\\/]*$

This thread has some examples for javascript.

Alternatively, you can so a string split on "\" to create an array and get the last one in the array.

edited May 23 '17 at 12:12

Community

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answered Nov 01 '12 at 11:10

Tony T

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Regex - get the fileName?

5 Answers5