How can I find all of the distinct file extensions in a folder hierarchy?

My awk-less, sed-less, Perl-less, Python-less POSIX-compliant alternative:

find . -type f | rev | cut -d. -f1 | rev  | tr '[:upper:]' '[:lower:]' | sort | uniq --count | sort -rn

The trick is that it reverses the line and cuts the extension at the beginning.
It also converts the extensions to lower case.

Example output:

   3689 jpg
   1036 png
    610 mp4
     90 webm
     90 mkv
     57 mov
     12 avi
     10 txt
      3 zip
      2 ogv
      1 xcf
      1 trashinfo
      1 sh
      1 m4v
      1 jpeg
      1 ini
      1 gqv
      1 gcs
      1 dv

Another way:

find . -type f -name "*.*" -printf "%f\n" | while IFS= read -r; do echo "${REPLY##*.}"; done | sort -u

You can drop the -name "*.*" but this ensures we are dealing only with files that do have an extension of some sort.

The -printf is find's print, not bash. -printf "%f\n" prints only the filename, stripping the path (and adds a newline).

Then we use string substitution to remove up to the last dot using ${REPLY##*.}.

Note that $REPLY is simply read's inbuilt variable. We could just as use our own in the form: while IFS= read -r file, and here $file would be the variable.

I don't think this one was mentioned yet:

find . -type f -exec sh -c 'echo "${0##*.}"' {} \; | sort | uniq -c

The accepted answer uses REGEX and you cannot create an alias command with REGEX, you have to put it into a shell script, I'm using Amazon Linux 2 and did the following:

1. I put the accepted answer code into a file using :

   sudo vim find.sh

add this code:

find ./ -type f | perl -ne 'print $1 if m/\.([^.\/]+)$/' | sort -u

save the file by typing: :wq!

2. sudo vim ~/.bash_profile

3. alias getext=". /path/to/your/find.sh"

4. :wq!

5. . ~/.bash_profile

I've found it simple and fast...

   # find . -type f -exec basename {} \; | awk -F"." '{print $NF}' > /tmp/outfile.txt
   # cat /tmp/outfile.txt | sort | uniq -c| sort -n > tmp/outfile_sorted.txt