How can I find all of the distinct file extensions

Question

On a Linux machine I would like to traverse a folder hierarchy and get a list of all of the distinct file extensions within it.

What would be the best way to achieve this from a shell?

Answer 1

Powershell:

dir -recurse | select-object extension -unique

Thanks to http://kevin-berridge.blogspot.com/2007/11/windows-powershell.html

Answer 2

I don't think this one was mentioned yet:

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

Answer 3

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.

Answer 4

Find everythin with a dot and show only the suffix.

find . -type f -name "*.*" | awk -F. '{print $NF}' | sort -u

if you know all suffix have 3 characters then

find . -type f -name "*.???" | awk -F. '{print $NF}' | sort -u

or with sed shows all suffixes with one to four characters. Change {1,4} to the range of characters you are expecting in the suffix.

find . -type f | sed -n 's/.*\.\(.\{1,4\}\)$/\1/p'| sort -u

Answer 5

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

Answer 6

you could also do this

find . -type f -name "*.php" -exec PATHTOAPP {} +