如何在文件夹层次结构中找到所有不同的文件扩展名？

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

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

无需管道sort ，awk 可以做到这一切：

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

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

Powershell:

dir -recurse | select-object extension -unique

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

Answer 5

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

Answer 6

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 7

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 -oE '\.(\w+)$' | sort -u

Answer 8

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 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.

Answer 9

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)

Answer 10

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)"

Answer 11

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

Answer 12

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 13

我认为还没有提到这个：

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

Answer 14

你也可以这样做

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

Answer 15

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

Answer 16

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

Answer 17

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.