lua get file extension first occurrence










1















So i have a url like so



https://example.com/path/to/file/file.mp4/file.jpg


I want to match only the first provided file extension provided not any others people insert into the url.



Example :



function GetFileExtension(url)
return url:match("^.+(%..+)$")
end

local url = "https://example.com/path/to/file/file.mp4/file.jpg"
print(GetFileExtension(url))


Output :



.jpg


The output should be a .mp4 since that is what the file is anything after the first occurrence is ignored on the url.



What is the best way to fix this. Thanks to anyone who can help me and answer my question.










share|improve this question

















  • 1





    url:match"//[^/]+/.-(%.[^/]+)"

    – Egor Skriptunoff
    Nov 14 '18 at 19:10






  • 1





    or (%.[^/]*) to allow empty extensions...

    – Paul Kulchenko
    Nov 14 '18 at 19:34











  • @PaulKulchenko - Are filenames like file. valid?

    – Egor Skriptunoff
    Nov 15 '18 at 7:28












  • Are URLs like https://example.com/path/to/file/../file.mp4/file.jpg possible?

    – Egor Skriptunoff
    Nov 15 '18 at 7:31















1















So i have a url like so



https://example.com/path/to/file/file.mp4/file.jpg


I want to match only the first provided file extension provided not any others people insert into the url.



Example :



function GetFileExtension(url)
return url:match("^.+(%..+)$")
end

local url = "https://example.com/path/to/file/file.mp4/file.jpg"
print(GetFileExtension(url))


Output :



.jpg


The output should be a .mp4 since that is what the file is anything after the first occurrence is ignored on the url.



What is the best way to fix this. Thanks to anyone who can help me and answer my question.










share|improve this question

















  • 1





    url:match"//[^/]+/.-(%.[^/]+)"

    – Egor Skriptunoff
    Nov 14 '18 at 19:10






  • 1





    or (%.[^/]*) to allow empty extensions...

    – Paul Kulchenko
    Nov 14 '18 at 19:34











  • @PaulKulchenko - Are filenames like file. valid?

    – Egor Skriptunoff
    Nov 15 '18 at 7:28












  • Are URLs like https://example.com/path/to/file/../file.mp4/file.jpg possible?

    – Egor Skriptunoff
    Nov 15 '18 at 7:31













1












1








1








So i have a url like so



https://example.com/path/to/file/file.mp4/file.jpg


I want to match only the first provided file extension provided not any others people insert into the url.



Example :



function GetFileExtension(url)
return url:match("^.+(%..+)$")
end

local url = "https://example.com/path/to/file/file.mp4/file.jpg"
print(GetFileExtension(url))


Output :



.jpg


The output should be a .mp4 since that is what the file is anything after the first occurrence is ignored on the url.



What is the best way to fix this. Thanks to anyone who can help me and answer my question.










share|improve this question














So i have a url like so



https://example.com/path/to/file/file.mp4/file.jpg


I want to match only the first provided file extension provided not any others people insert into the url.



Example :



function GetFileExtension(url)
return url:match("^.+(%..+)$")
end

local url = "https://example.com/path/to/file/file.mp4/file.jpg"
print(GetFileExtension(url))


Output :



.jpg


The output should be a .mp4 since that is what the file is anything after the first occurrence is ignored on the url.



What is the best way to fix this. Thanks to anyone who can help me and answer my question.







lua






share|improve this question













share|improve this question











share|improve this question




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asked Nov 14 '18 at 18:48









C0nw0nkC0nw0nk

3721214




3721214







  • 1





    url:match"//[^/]+/.-(%.[^/]+)"

    – Egor Skriptunoff
    Nov 14 '18 at 19:10






  • 1





    or (%.[^/]*) to allow empty extensions...

    – Paul Kulchenko
    Nov 14 '18 at 19:34











  • @PaulKulchenko - Are filenames like file. valid?

    – Egor Skriptunoff
    Nov 15 '18 at 7:28












  • Are URLs like https://example.com/path/to/file/../file.mp4/file.jpg possible?

    – Egor Skriptunoff
    Nov 15 '18 at 7:31












  • 1





    url:match"//[^/]+/.-(%.[^/]+)"

    – Egor Skriptunoff
    Nov 14 '18 at 19:10






  • 1





    or (%.[^/]*) to allow empty extensions...

    – Paul Kulchenko
    Nov 14 '18 at 19:34











  • @PaulKulchenko - Are filenames like file. valid?

    – Egor Skriptunoff
    Nov 15 '18 at 7:28












  • Are URLs like https://example.com/path/to/file/../file.mp4/file.jpg possible?

    – Egor Skriptunoff
    Nov 15 '18 at 7:31







1




1





url:match"//[^/]+/.-(%.[^/]+)"

– Egor Skriptunoff
Nov 14 '18 at 19:10





url:match"//[^/]+/.-(%.[^/]+)"

– Egor Skriptunoff
Nov 14 '18 at 19:10




1




1





or (%.[^/]*) to allow empty extensions...

– Paul Kulchenko
Nov 14 '18 at 19:34





or (%.[^/]*) to allow empty extensions...

– Paul Kulchenko
Nov 14 '18 at 19:34













@PaulKulchenko - Are filenames like file. valid?

– Egor Skriptunoff
Nov 15 '18 at 7:28






@PaulKulchenko - Are filenames like file. valid?

– Egor Skriptunoff
Nov 15 '18 at 7:28














Are URLs like https://example.com/path/to/file/../file.mp4/file.jpg possible?

– Egor Skriptunoff
Nov 15 '18 at 7:31





Are URLs like https://example.com/path/to/file/../file.mp4/file.jpg possible?

– Egor Skriptunoff
Nov 15 '18 at 7:31












1 Answer
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oldest

votes


















0














Try url:match("//.-/.+(%..*)$")).



The pattern finds the first / after //, thus skipping the host part. Then it finds the last . and captures it together with the extension, if any.






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    1 Answer
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    0














    Try url:match("//.-/.+(%..*)$")).



    The pattern finds the first / after //, thus skipping the host part. Then it finds the last . and captures it together with the extension, if any.






    share|improve this answer



























      0














      Try url:match("//.-/.+(%..*)$")).



      The pattern finds the first / after //, thus skipping the host part. Then it finds the last . and captures it together with the extension, if any.






      share|improve this answer

























        0












        0








        0







        Try url:match("//.-/.+(%..*)$")).



        The pattern finds the first / after //, thus skipping the host part. Then it finds the last . and captures it together with the extension, if any.






        share|improve this answer













        Try url:match("//.-/.+(%..*)$")).



        The pattern finds the first / after //, thus skipping the host part. Then it finds the last . and captures it together with the extension, if any.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 15 '18 at 8:59









        lhflhf

        56.2k668104




        56.2k668104





























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