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

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

Answer 1

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 2

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 3

you could also do this

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

Answer 4

Adding my own variation to the mix. I think it's the simplest of the lot and can be useful when efficiency is not a big concern.

find . -type f | grep -o -E '\.[^\.]+$' | sort -u

Answer 5

I think the most simple & straightforward way is

for f in *.*; do echo "${f##*.}"; done | sort -u

It's modified on ChristopheD's 3rd way.

Answer 6

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