linux find command for filenames without extension for unknown extensions












3















I like using the following command to print file names but it prints them with extensions.



find . -type f -printf '%fn'


In my directory, there are many files with different extensions. I tried adding --ignore='*.*' both before and after -printf, but it didn't work.
example; I have files myfile1.txt, myfile2.mp3, etc. I need it prints myfile1, myfile2, etc.
How would I do this?










share|improve this question

























  • Since you specifically ask about find I’ll provide this aside in a comment. You can do this fluently with modern native bash: shopt -s globstar; files=(**/*); printf '%sn' "${files[@]%.*}" — this could have odd behavior if a directory in the path has a . in it and the files in that path have no ., but I assume that combination is unlikely.

    – kojiro
    2 days ago
















3















I like using the following command to print file names but it prints them with extensions.



find . -type f -printf '%fn'


In my directory, there are many files with different extensions. I tried adding --ignore='*.*' both before and after -printf, but it didn't work.
example; I have files myfile1.txt, myfile2.mp3, etc. I need it prints myfile1, myfile2, etc.
How would I do this?










share|improve this question

























  • Since you specifically ask about find I’ll provide this aside in a comment. You can do this fluently with modern native bash: shopt -s globstar; files=(**/*); printf '%sn' "${files[@]%.*}" — this could have odd behavior if a directory in the path has a . in it and the files in that path have no ., but I assume that combination is unlikely.

    – kojiro
    2 days ago














3












3








3








I like using the following command to print file names but it prints them with extensions.



find . -type f -printf '%fn'


In my directory, there are many files with different extensions. I tried adding --ignore='*.*' both before and after -printf, but it didn't work.
example; I have files myfile1.txt, myfile2.mp3, etc. I need it prints myfile1, myfile2, etc.
How would I do this?










share|improve this question
















I like using the following command to print file names but it prints them with extensions.



find . -type f -printf '%fn'


In my directory, there are many files with different extensions. I tried adding --ignore='*.*' both before and after -printf, but it didn't work.
example; I have files myfile1.txt, myfile2.mp3, etc. I need it prints myfile1, myfile2, etc.
How would I do this?







find filenames






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 days ago







kutlus

















asked 2 days ago









kutluskutlus

485




485













  • Since you specifically ask about find I’ll provide this aside in a comment. You can do this fluently with modern native bash: shopt -s globstar; files=(**/*); printf '%sn' "${files[@]%.*}" — this could have odd behavior if a directory in the path has a . in it and the files in that path have no ., but I assume that combination is unlikely.

    – kojiro
    2 days ago



















  • Since you specifically ask about find I’ll provide this aside in a comment. You can do this fluently with modern native bash: shopt -s globstar; files=(**/*); printf '%sn' "${files[@]%.*}" — this could have odd behavior if a directory in the path has a . in it and the files in that path have no ., but I assume that combination is unlikely.

    – kojiro
    2 days ago

















Since you specifically ask about find I’ll provide this aside in a comment. You can do this fluently with modern native bash: shopt -s globstar; files=(**/*); printf '%sn' "${files[@]%.*}" — this could have odd behavior if a directory in the path has a . in it and the files in that path have no ., but I assume that combination is unlikely.

– kojiro
2 days ago





Since you specifically ask about find I’ll provide this aside in a comment. You can do this fluently with modern native bash: shopt -s globstar; files=(**/*); printf '%sn' "${files[@]%.*}" — this could have odd behavior if a directory in the path has a . in it and the files in that path have no ., but I assume that combination is unlikely.

– kojiro
2 days ago










4 Answers
4






active

oldest

votes


















3














If I understand you correctly (I didn't, see second part of the answer), you want to avoid listing filenames that contain a dot character.



This will do that:



find . -type f ! -name '*.*'


The filename globbing pattern *.* would match any filename containing at least one dot. The preceding ! negates the sense of the match, which means that the pathnames that gets through to the end will be those of files that have no dots in their names. In really ancient shells, you may want to escape the ! as ! (or update your Unix installation). The lone ! won't invoke bash's history expansion facility.



To print only the filename component of the found pathname, with GNU find:



find . -type f ! -name '*.*' -printf '%fn'


With standard find (or GNU find for that matter):



find . -type f ! -name '*.*' -exec basename {} ;


Before using this in a command substitution in a loop, see "Why is looping over find's output bad practice?".





To list all filenames, and at the same time remove everything after the last dot in the name ("remove the extension"), you may use



find . -type f -exec sh -c '
for pathname do
pathname=$( basename "$pathname" )
printf "%sn" "${pathname%.*}"
done' sh {} +


This would send all found pathnames of all files to a short shell loop. The loop would take each pathname and call basename on it to extract the filename component of the pathname, and then print the resulting string with everything after the last dot removed.



The parameter expansion ${pathname%.*} means "remove the shortest string matching .* (a literal dot followed by arbitrary text) from the end of the value of $pathname". It would have the effect of removing a filename suffix after the last dot in the filename.



For more info about find ... -exec ... {} +, see e.g. "Understanding the -exec option of `find`".






share|improve this answer


























  • Thanks for your help, I have tried all three, but they don't print anything. Am I missing something, on the same terminal, in the same directory find . -type f -printf '%fn' works but as I asked in the question I need them without extensions.

    – kutlus
    2 days ago











  • @kutlus In that case, I (and the others) may have misunderstood your question. Do you want to remove everything after the last dot in the filename?

    – Kusalananda
    2 days ago











  • I just want the file name, example; some files are myfile1.txt, myfile2.mp3, i want it prints myfile1, myfile2. sorry if my questions wasn`t clear, thanks

    – kutlus
    2 days ago











  • @kutlus See updated answer.

    – Kusalananda
    2 days ago











  • This worked, many thanks!

    – kutlus
    2 days ago



















3














In Linux, there is no such thing as a file extension. A . in a file name has no significance whatsoever (notwithstanding that a . as the first character in a filename identifies it as a hidden file).



Also, checking the manual page for find on my system shows no --ignore option.



That said, if you want to ignore any files with a . in their name, you can use find's -not operator:



find -type f -not -name '*.*' -print





share|improve this answer
























  • thanks but this doesn`t print anything.

    – kutlus
    2 days ago











  • If this does not print anything, then all files must have a . in their names.

    – DopeGhoti
    2 days ago



















2














try



find . -type f ! -name '*.*' -print


where





  • ! : not (! must be escaped)


  • name '*.*' : filename with extension






share|improve this answer
























  • arg, not fast enough !!

    – Archemar
    2 days ago











  • Having to escape ! is why I usually suggest using -not.

    – DopeGhoti
    2 days ago











  • well it work w/o escape, I've been escaping it ever since SunOS 3.X... sight.

    – Archemar
    2 days ago











  • thanks but this doesn`t print anything.

    – kutlus
    2 days ago











  • In what shell do you need to escape it? It works without escaping in both zsh and bash. Both these shells only interpret a history expansion when the ! is accompanied by another character.

    – JoL
    2 days ago



















2














As other's have pointed out, "extensions" don't really mean anything to Unix systems. But we can remove them from a listing with a simple sed command.



e.g.



find * -type f -print | sed 's/.[^.]*$//'


If there are directories and you don't want them to be shown in the listing then



find * -type f -printf "%fn" | sed 's/.[^.]*$//'


For a single directory we can just use ls instead of find



ls | sed 's/.[^.]*$//'





share|improve this answer























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    4 Answers
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    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    If I understand you correctly (I didn't, see second part of the answer), you want to avoid listing filenames that contain a dot character.



    This will do that:



    find . -type f ! -name '*.*'


    The filename globbing pattern *.* would match any filename containing at least one dot. The preceding ! negates the sense of the match, which means that the pathnames that gets through to the end will be those of files that have no dots in their names. In really ancient shells, you may want to escape the ! as ! (or update your Unix installation). The lone ! won't invoke bash's history expansion facility.



    To print only the filename component of the found pathname, with GNU find:



    find . -type f ! -name '*.*' -printf '%fn'


    With standard find (or GNU find for that matter):



    find . -type f ! -name '*.*' -exec basename {} ;


    Before using this in a command substitution in a loop, see "Why is looping over find's output bad practice?".





    To list all filenames, and at the same time remove everything after the last dot in the name ("remove the extension"), you may use



    find . -type f -exec sh -c '
    for pathname do
    pathname=$( basename "$pathname" )
    printf "%sn" "${pathname%.*}"
    done' sh {} +


    This would send all found pathnames of all files to a short shell loop. The loop would take each pathname and call basename on it to extract the filename component of the pathname, and then print the resulting string with everything after the last dot removed.



    The parameter expansion ${pathname%.*} means "remove the shortest string matching .* (a literal dot followed by arbitrary text) from the end of the value of $pathname". It would have the effect of removing a filename suffix after the last dot in the filename.



    For more info about find ... -exec ... {} +, see e.g. "Understanding the -exec option of `find`".






    share|improve this answer


























    • Thanks for your help, I have tried all three, but they don't print anything. Am I missing something, on the same terminal, in the same directory find . -type f -printf '%fn' works but as I asked in the question I need them without extensions.

      – kutlus
      2 days ago











    • @kutlus In that case, I (and the others) may have misunderstood your question. Do you want to remove everything after the last dot in the filename?

      – Kusalananda
      2 days ago











    • I just want the file name, example; some files are myfile1.txt, myfile2.mp3, i want it prints myfile1, myfile2. sorry if my questions wasn`t clear, thanks

      – kutlus
      2 days ago











    • @kutlus See updated answer.

      – Kusalananda
      2 days ago











    • This worked, many thanks!

      – kutlus
      2 days ago
















    3














    If I understand you correctly (I didn't, see second part of the answer), you want to avoid listing filenames that contain a dot character.



    This will do that:



    find . -type f ! -name '*.*'


    The filename globbing pattern *.* would match any filename containing at least one dot. The preceding ! negates the sense of the match, which means that the pathnames that gets through to the end will be those of files that have no dots in their names. In really ancient shells, you may want to escape the ! as ! (or update your Unix installation). The lone ! won't invoke bash's history expansion facility.



    To print only the filename component of the found pathname, with GNU find:



    find . -type f ! -name '*.*' -printf '%fn'


    With standard find (or GNU find for that matter):



    find . -type f ! -name '*.*' -exec basename {} ;


    Before using this in a command substitution in a loop, see "Why is looping over find's output bad practice?".





    To list all filenames, and at the same time remove everything after the last dot in the name ("remove the extension"), you may use



    find . -type f -exec sh -c '
    for pathname do
    pathname=$( basename "$pathname" )
    printf "%sn" "${pathname%.*}"
    done' sh {} +


    This would send all found pathnames of all files to a short shell loop. The loop would take each pathname and call basename on it to extract the filename component of the pathname, and then print the resulting string with everything after the last dot removed.



    The parameter expansion ${pathname%.*} means "remove the shortest string matching .* (a literal dot followed by arbitrary text) from the end of the value of $pathname". It would have the effect of removing a filename suffix after the last dot in the filename.



    For more info about find ... -exec ... {} +, see e.g. "Understanding the -exec option of `find`".






    share|improve this answer


























    • Thanks for your help, I have tried all three, but they don't print anything. Am I missing something, on the same terminal, in the same directory find . -type f -printf '%fn' works but as I asked in the question I need them without extensions.

      – kutlus
      2 days ago











    • @kutlus In that case, I (and the others) may have misunderstood your question. Do you want to remove everything after the last dot in the filename?

      – Kusalananda
      2 days ago











    • I just want the file name, example; some files are myfile1.txt, myfile2.mp3, i want it prints myfile1, myfile2. sorry if my questions wasn`t clear, thanks

      – kutlus
      2 days ago











    • @kutlus See updated answer.

      – Kusalananda
      2 days ago











    • This worked, many thanks!

      – kutlus
      2 days ago














    3












    3








    3







    If I understand you correctly (I didn't, see second part of the answer), you want to avoid listing filenames that contain a dot character.



    This will do that:



    find . -type f ! -name '*.*'


    The filename globbing pattern *.* would match any filename containing at least one dot. The preceding ! negates the sense of the match, which means that the pathnames that gets through to the end will be those of files that have no dots in their names. In really ancient shells, you may want to escape the ! as ! (or update your Unix installation). The lone ! won't invoke bash's history expansion facility.



    To print only the filename component of the found pathname, with GNU find:



    find . -type f ! -name '*.*' -printf '%fn'


    With standard find (or GNU find for that matter):



    find . -type f ! -name '*.*' -exec basename {} ;


    Before using this in a command substitution in a loop, see "Why is looping over find's output bad practice?".





    To list all filenames, and at the same time remove everything after the last dot in the name ("remove the extension"), you may use



    find . -type f -exec sh -c '
    for pathname do
    pathname=$( basename "$pathname" )
    printf "%sn" "${pathname%.*}"
    done' sh {} +


    This would send all found pathnames of all files to a short shell loop. The loop would take each pathname and call basename on it to extract the filename component of the pathname, and then print the resulting string with everything after the last dot removed.



    The parameter expansion ${pathname%.*} means "remove the shortest string matching .* (a literal dot followed by arbitrary text) from the end of the value of $pathname". It would have the effect of removing a filename suffix after the last dot in the filename.



    For more info about find ... -exec ... {} +, see e.g. "Understanding the -exec option of `find`".






    share|improve this answer















    If I understand you correctly (I didn't, see second part of the answer), you want to avoid listing filenames that contain a dot character.



    This will do that:



    find . -type f ! -name '*.*'


    The filename globbing pattern *.* would match any filename containing at least one dot. The preceding ! negates the sense of the match, which means that the pathnames that gets through to the end will be those of files that have no dots in their names. In really ancient shells, you may want to escape the ! as ! (or update your Unix installation). The lone ! won't invoke bash's history expansion facility.



    To print only the filename component of the found pathname, with GNU find:



    find . -type f ! -name '*.*' -printf '%fn'


    With standard find (or GNU find for that matter):



    find . -type f ! -name '*.*' -exec basename {} ;


    Before using this in a command substitution in a loop, see "Why is looping over find's output bad practice?".





    To list all filenames, and at the same time remove everything after the last dot in the name ("remove the extension"), you may use



    find . -type f -exec sh -c '
    for pathname do
    pathname=$( basename "$pathname" )
    printf "%sn" "${pathname%.*}"
    done' sh {} +


    This would send all found pathnames of all files to a short shell loop. The loop would take each pathname and call basename on it to extract the filename component of the pathname, and then print the resulting string with everything after the last dot removed.



    The parameter expansion ${pathname%.*} means "remove the shortest string matching .* (a literal dot followed by arbitrary text) from the end of the value of $pathname". It would have the effect of removing a filename suffix after the last dot in the filename.



    For more info about find ... -exec ... {} +, see e.g. "Understanding the -exec option of `find`".







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 2 days ago

























    answered 2 days ago









    KusalanandaKusalananda

    126k16239393




    126k16239393













    • Thanks for your help, I have tried all three, but they don't print anything. Am I missing something, on the same terminal, in the same directory find . -type f -printf '%fn' works but as I asked in the question I need them without extensions.

      – kutlus
      2 days ago











    • @kutlus In that case, I (and the others) may have misunderstood your question. Do you want to remove everything after the last dot in the filename?

      – Kusalananda
      2 days ago











    • I just want the file name, example; some files are myfile1.txt, myfile2.mp3, i want it prints myfile1, myfile2. sorry if my questions wasn`t clear, thanks

      – kutlus
      2 days ago











    • @kutlus See updated answer.

      – Kusalananda
      2 days ago











    • This worked, many thanks!

      – kutlus
      2 days ago



















    • Thanks for your help, I have tried all three, but they don't print anything. Am I missing something, on the same terminal, in the same directory find . -type f -printf '%fn' works but as I asked in the question I need them without extensions.

      – kutlus
      2 days ago











    • @kutlus In that case, I (and the others) may have misunderstood your question. Do you want to remove everything after the last dot in the filename?

      – Kusalananda
      2 days ago











    • I just want the file name, example; some files are myfile1.txt, myfile2.mp3, i want it prints myfile1, myfile2. sorry if my questions wasn`t clear, thanks

      – kutlus
      2 days ago











    • @kutlus See updated answer.

      – Kusalananda
      2 days ago











    • This worked, many thanks!

      – kutlus
      2 days ago

















    Thanks for your help, I have tried all three, but they don't print anything. Am I missing something, on the same terminal, in the same directory find . -type f -printf '%fn' works but as I asked in the question I need them without extensions.

    – kutlus
    2 days ago





    Thanks for your help, I have tried all three, but they don't print anything. Am I missing something, on the same terminal, in the same directory find . -type f -printf '%fn' works but as I asked in the question I need them without extensions.

    – kutlus
    2 days ago













    @kutlus In that case, I (and the others) may have misunderstood your question. Do you want to remove everything after the last dot in the filename?

    – Kusalananda
    2 days ago





    @kutlus In that case, I (and the others) may have misunderstood your question. Do you want to remove everything after the last dot in the filename?

    – Kusalananda
    2 days ago













    I just want the file name, example; some files are myfile1.txt, myfile2.mp3, i want it prints myfile1, myfile2. sorry if my questions wasn`t clear, thanks

    – kutlus
    2 days ago





    I just want the file name, example; some files are myfile1.txt, myfile2.mp3, i want it prints myfile1, myfile2. sorry if my questions wasn`t clear, thanks

    – kutlus
    2 days ago













    @kutlus See updated answer.

    – Kusalananda
    2 days ago





    @kutlus See updated answer.

    – Kusalananda
    2 days ago













    This worked, many thanks!

    – kutlus
    2 days ago





    This worked, many thanks!

    – kutlus
    2 days ago













    3














    In Linux, there is no such thing as a file extension. A . in a file name has no significance whatsoever (notwithstanding that a . as the first character in a filename identifies it as a hidden file).



    Also, checking the manual page for find on my system shows no --ignore option.



    That said, if you want to ignore any files with a . in their name, you can use find's -not operator:



    find -type f -not -name '*.*' -print





    share|improve this answer
























    • thanks but this doesn`t print anything.

      – kutlus
      2 days ago











    • If this does not print anything, then all files must have a . in their names.

      – DopeGhoti
      2 days ago
















    3














    In Linux, there is no such thing as a file extension. A . in a file name has no significance whatsoever (notwithstanding that a . as the first character in a filename identifies it as a hidden file).



    Also, checking the manual page for find on my system shows no --ignore option.



    That said, if you want to ignore any files with a . in their name, you can use find's -not operator:



    find -type f -not -name '*.*' -print





    share|improve this answer
























    • thanks but this doesn`t print anything.

      – kutlus
      2 days ago











    • If this does not print anything, then all files must have a . in their names.

      – DopeGhoti
      2 days ago














    3












    3








    3







    In Linux, there is no such thing as a file extension. A . in a file name has no significance whatsoever (notwithstanding that a . as the first character in a filename identifies it as a hidden file).



    Also, checking the manual page for find on my system shows no --ignore option.



    That said, if you want to ignore any files with a . in their name, you can use find's -not operator:



    find -type f -not -name '*.*' -print





    share|improve this answer













    In Linux, there is no such thing as a file extension. A . in a file name has no significance whatsoever (notwithstanding that a . as the first character in a filename identifies it as a hidden file).



    Also, checking the manual page for find on my system shows no --ignore option.



    That said, if you want to ignore any files with a . in their name, you can use find's -not operator:



    find -type f -not -name '*.*' -print






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 2 days ago









    DopeGhotiDopeGhoti

    44.4k55684




    44.4k55684













    • thanks but this doesn`t print anything.

      – kutlus
      2 days ago











    • If this does not print anything, then all files must have a . in their names.

      – DopeGhoti
      2 days ago



















    • thanks but this doesn`t print anything.

      – kutlus
      2 days ago











    • If this does not print anything, then all files must have a . in their names.

      – DopeGhoti
      2 days ago

















    thanks but this doesn`t print anything.

    – kutlus
    2 days ago





    thanks but this doesn`t print anything.

    – kutlus
    2 days ago













    If this does not print anything, then all files must have a . in their names.

    – DopeGhoti
    2 days ago





    If this does not print anything, then all files must have a . in their names.

    – DopeGhoti
    2 days ago











    2














    try



    find . -type f ! -name '*.*' -print


    where





    • ! : not (! must be escaped)


    • name '*.*' : filename with extension






    share|improve this answer
























    • arg, not fast enough !!

      – Archemar
      2 days ago











    • Having to escape ! is why I usually suggest using -not.

      – DopeGhoti
      2 days ago











    • well it work w/o escape, I've been escaping it ever since SunOS 3.X... sight.

      – Archemar
      2 days ago











    • thanks but this doesn`t print anything.

      – kutlus
      2 days ago











    • In what shell do you need to escape it? It works without escaping in both zsh and bash. Both these shells only interpret a history expansion when the ! is accompanied by another character.

      – JoL
      2 days ago
















    2














    try



    find . -type f ! -name '*.*' -print


    where





    • ! : not (! must be escaped)


    • name '*.*' : filename with extension






    share|improve this answer
























    • arg, not fast enough !!

      – Archemar
      2 days ago











    • Having to escape ! is why I usually suggest using -not.

      – DopeGhoti
      2 days ago











    • well it work w/o escape, I've been escaping it ever since SunOS 3.X... sight.

      – Archemar
      2 days ago











    • thanks but this doesn`t print anything.

      – kutlus
      2 days ago











    • In what shell do you need to escape it? It works without escaping in both zsh and bash. Both these shells only interpret a history expansion when the ! is accompanied by another character.

      – JoL
      2 days ago














    2












    2








    2







    try



    find . -type f ! -name '*.*' -print


    where





    • ! : not (! must be escaped)


    • name '*.*' : filename with extension






    share|improve this answer













    try



    find . -type f ! -name '*.*' -print


    where





    • ! : not (! must be escaped)


    • name '*.*' : filename with extension







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 2 days ago









    ArchemarArchemar

    19.9k93670




    19.9k93670













    • arg, not fast enough !!

      – Archemar
      2 days ago











    • Having to escape ! is why I usually suggest using -not.

      – DopeGhoti
      2 days ago











    • well it work w/o escape, I've been escaping it ever since SunOS 3.X... sight.

      – Archemar
      2 days ago











    • thanks but this doesn`t print anything.

      – kutlus
      2 days ago











    • In what shell do you need to escape it? It works without escaping in both zsh and bash. Both these shells only interpret a history expansion when the ! is accompanied by another character.

      – JoL
      2 days ago



















    • arg, not fast enough !!

      – Archemar
      2 days ago











    • Having to escape ! is why I usually suggest using -not.

      – DopeGhoti
      2 days ago











    • well it work w/o escape, I've been escaping it ever since SunOS 3.X... sight.

      – Archemar
      2 days ago











    • thanks but this doesn`t print anything.

      – kutlus
      2 days ago











    • In what shell do you need to escape it? It works without escaping in both zsh and bash. Both these shells only interpret a history expansion when the ! is accompanied by another character.

      – JoL
      2 days ago

















    arg, not fast enough !!

    – Archemar
    2 days ago





    arg, not fast enough !!

    – Archemar
    2 days ago













    Having to escape ! is why I usually suggest using -not.

    – DopeGhoti
    2 days ago





    Having to escape ! is why I usually suggest using -not.

    – DopeGhoti
    2 days ago













    well it work w/o escape, I've been escaping it ever since SunOS 3.X... sight.

    – Archemar
    2 days ago





    well it work w/o escape, I've been escaping it ever since SunOS 3.X... sight.

    – Archemar
    2 days ago













    thanks but this doesn`t print anything.

    – kutlus
    2 days ago





    thanks but this doesn`t print anything.

    – kutlus
    2 days ago













    In what shell do you need to escape it? It works without escaping in both zsh and bash. Both these shells only interpret a history expansion when the ! is accompanied by another character.

    – JoL
    2 days ago





    In what shell do you need to escape it? It works without escaping in both zsh and bash. Both these shells only interpret a history expansion when the ! is accompanied by another character.

    – JoL
    2 days ago











    2














    As other's have pointed out, "extensions" don't really mean anything to Unix systems. But we can remove them from a listing with a simple sed command.



    e.g.



    find * -type f -print | sed 's/.[^.]*$//'


    If there are directories and you don't want them to be shown in the listing then



    find * -type f -printf "%fn" | sed 's/.[^.]*$//'


    For a single directory we can just use ls instead of find



    ls | sed 's/.[^.]*$//'





    share|improve this answer




























      2














      As other's have pointed out, "extensions" don't really mean anything to Unix systems. But we can remove them from a listing with a simple sed command.



      e.g.



      find * -type f -print | sed 's/.[^.]*$//'


      If there are directories and you don't want them to be shown in the listing then



      find * -type f -printf "%fn" | sed 's/.[^.]*$//'


      For a single directory we can just use ls instead of find



      ls | sed 's/.[^.]*$//'





      share|improve this answer


























        2












        2








        2







        As other's have pointed out, "extensions" don't really mean anything to Unix systems. But we can remove them from a listing with a simple sed command.



        e.g.



        find * -type f -print | sed 's/.[^.]*$//'


        If there are directories and you don't want them to be shown in the listing then



        find * -type f -printf "%fn" | sed 's/.[^.]*$//'


        For a single directory we can just use ls instead of find



        ls | sed 's/.[^.]*$//'





        share|improve this answer













        As other's have pointed out, "extensions" don't really mean anything to Unix systems. But we can remove them from a listing with a simple sed command.



        e.g.



        find * -type f -print | sed 's/.[^.]*$//'


        If there are directories and you don't want them to be shown in the listing then



        find * -type f -printf "%fn" | sed 's/.[^.]*$//'


        For a single directory we can just use ls instead of find



        ls | sed 's/.[^.]*$//'






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 2 days ago









        Stephen HarrisStephen Harris

        25.5k24477




        25.5k24477






























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