How can I find all of the distinct file extensions

问题:

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?

回答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

回答2:

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

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

回答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.

回答4:

Powershell:

dir -recurse | select-object extension -unique

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

回答5:

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

回答6:

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

回答7:

I tried a bunch of the answers here, even the "best" answer. They all came up short of what I specifically was after. So besides the past 12 hours of sitting in regex code for multiple programs and reading and testing these answers this is what I came up with which works EXACTLY like I want.

 find . -type f -name "*.*" | grep -o -E "\.[^\.]+$" | grep -o -E "[[:alpha:]]{2,16}" | awk '{print tolower($0)}' | sort -u

• Finds all files which may have an extension.
• Greps only the extension
• Greps for file extensions between 2 and 16 characters (just adjust the numbers if they don't fit your need). This helps avoid cache files and system files (system file bit is to search jail).
• Awk to print the extensions in lower case.
• Sort and bring in only unique values. Originally I had attempted to try the awk answer but it would double print items that varied in case sensitivity.

If you need a count of the file extensions then use the below code

find . -type f -name "*.*" | grep -o -E "\.[^\.]+$" | grep -o -E "[[:alpha:]]{2,16}" | awk '{print tolower($0)}' | sort | uniq -c | sort -rn

While these methods will take some time to complete and probably aren't the the best ways to go about the problem, they work.

Update: Per @alpha_989 long file extensions will cause an issue. That's due to the original regex "[[:alpha:]]{3,6}". I have updated the answer to include the regex "[[:alpha:]]{2,16}". However anyone using this code should be aware that those numbers are the min and max of how long the extension is allowed for the final output. Anything outside that range will be split into multiple lines in the output.

Note: Original post did read "- Greps for file extensions between 3 and 6 characters (just adjust the numbers if they don't fit your need). This helps avoid cache files and system files (system file bit is to search jail)."

Idea: Could be used to find file extensions over a specific length via:

 find . -type f -name "*.*" | grep -o -E "\.[^\.]+$" | grep -o -E "[[:alpha:]]{4,}" | awk '{print tolower($0)}' | sort -u

Where 4 is the file extensions length to include and then find also any extensions beyond that length.

回答8:

In Python using generators for very large directories, including blank extensions, and getting the number of times each extension shows up:

import json
import collections
import itertools
import os

root = '/home/andres'
files = itertools.chain.from_iterable((
    files for _,_,files in os.walk(root)
    ))
counter = collections.Counter(
    (os.path.splitext(file_)[1] for file_ in files)
)
print json.dumps(counter, indent=2)

回答9:

Since there's already another solution which uses Perl:

If you have Python installed you could also do (from the shell):

python -c "import os;e=set();[[e.add(os.path.splitext(f)[-1]) for f in fn]for _,_,fn in os.walk('/home')];print '\n'.join(e)"

回答10:

None of the replies so far deal with filenames with newlines properly (except for ChristopheD's, which just came in as I was typing this). The following is not a shell one-liner, but works, and is reasonably fast.

import os, sys

def names(roots):
    for root in roots:
        for a, b, basenames in os.walk(root):
            for basename in basenames:
                yield basename

sufs = set(os.path.splitext(x)[1] for x in names(sys.argv[1:]))
for suf in sufs:
    if suf:
        print suf

回答11:

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.

回答12:

I don't think this one was mentioned yet:

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

回答13:

you could also do this

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