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

Question

Try this (not sure if it's the best way, but it works):

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

It work as following:

• Find all files from current folder
• Prints extension of files if any
• Make a unique sorted list

No need for the pipe to sort, awk can do it all:

find . -type f | awk -F. '!a[$NF]++{print $NF}'

Recursive version:

find . -type f | sed -e 's/.*\.//' | sed -e 's/.*\///' | sort -u

If you want totals (how may times the extension was seen):

find . -type f | sed -e 's/.*\.//' | sed -e 's/.*\///' | sort | uniq -c | sort -rn

Non-recursive (single folder):

for f in *.*; do printf "%s\n" "${f##*.}"; done | sort -u

I've based this upon this forum post, credit should go there.

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