Using 'find' to return filenames without extension

Question

I have a directory (with subdirectories), of which I want to find all files that have a ".ipynb" extension. But I want the 'find' command to just return me these filenames without the extension.

I know the first part:

find . -type f -iname "*.ipynb" -print    

But how do I then get the names without the "ipynb" extension? Any replies greatly appreciated...

Answer 1

To return only filenames without the extension, try:

find . -type f -iname "*.ipynb" -execdir sh -c 'printf "%s\n" "${0%.*}"' {} ';'

or (omitting -type f from now on):

find "$PWD" -iname "*.ipynb" -execdir basename {} .ipynb ';'

or:

find . -iname "*.ipynb" -exec basename {} .ipynb ';'

or:

find . -iname "*.ipynb" | sed "s/.*\///; s/\.ipynb//"

however invoking basename on each file can be inefficient, so @CharlesDuffy suggestion is:

find . -iname '*.ipynb' -exec bash -c 'printf "%s\n" "${@%.*}"' _ {} +

or:

find . -iname '*.ipynb' -execdir basename -s '.sh' {} +

^{Using + means that we're passing multiple files to each bash instance, so if the whole list fits into a single command line, we call bash only once.}

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To print full path and filename (without extension) in the same line, try:

find . -iname "*.ipynb" -exec sh -c 'printf "%s\n" "${0%.*}"' {} ';'

or:

find "$PWD" -iname "*.ipynb" -print | grep -o "[^\.]\+"

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To print full path and filename on separate lines:

find "$PWD" -iname "*.ipynb" -exec dirname "{}" ';' -exec basename "{}" .ipynb ';'

Answer 2

Here's a simple solution:

find . -type f -iname "*.ipynb" | sed 's/\.ipynb$//1'

Answer 3

I found this in a bash oneliner that simplifies the process without using find

for n in *.ipynb; do echo "${n%.ipynb}"; done

Answer 4

If you need to have the name with directory but without the extension :

find .  -type f -iname "*.ipynb" -exec sh -c 'f=$(basename $1 .ipynb);d=$(dirname $1);echo "$d/$f"' sh {} \;