blocks|key|1390676|text|不需要到sort的管道，awk可以做所有的事情：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1390677|find+.+-type+f+%7C+awk+-F.+'!a[$NF]%2B%2B{print+$NF}'|code-block|syntax|javascript|1390678|entityMap^0|4|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

No need for the pipe to <code>sort</code>, awk can do it all: 

<pre><code>find . -type f | awk -F. '!a[$NF]++{print $NF}'
</code></pre>

blocks|key|1390456|text|递归版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1390457|find+.+-type+f+%7C+sed+-e+'s/.*\.//'+%7C+sed+-e+'s/.*\///'+%7C+sort+-u|code-block|syntax|javascript|1390458|如果您想要总计(如何查看扩展的时间)：|1390459|find+.+-type+f+%7C+sed+-e+'s/.*\.//'+%7C+sed+-e+'s/.*\///'+%7C+sort+%7C+uniq+-c+%7C+sort+-rn|1390460|非递归(单个文件夹)：|1390461|for+f+in+*.*;+do+printf+"%25s\n"+"${f##*.}";+done+%7C+sort+-u|1390462|我是基于this+forum+post的，功劳应该放在那里。|offset|length|1390463|entityMap|0|LINK|mutability|MUTABLE|url|http://www.unix.com/shell-programming-scripting/60471-recursicely-search-rename-file-extension.html#post302183807^0|0|0|0|0|0|0|4|F|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|16|8|@]|9|@$Q|17|R|18|1|19]]|A|$]]|$1|S|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

Recursive version:

<pre><code>find . -type f | sed -e 's/.*\.//' | sed -e 's/.*\///' | sort -u
</code></pre>

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

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

Non-recursive (single folder):

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

I've based this upon <a href="http://www.unix.com/shell-programming-scripting/60471-recursicely-search-rename-file-extension.html#post302183807" rel="noreferrer">this forum post</a>, credit should go there.

blocks|key|1390637|text|Powershell：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1390638|dir+-recurse+%7C+select-object+extension+-unique|code-block|syntax|javascript|1390639|感谢http://kevin-berridge.blogspot.com/2007/11/windows-powershell.html|offset|length|1390640|entityMap|0|LINK|mutability|MUTABLE|url|http://kevin-berridge.blogspot.com/2007/11/windows-powershell.html^0|0|0|2|1U|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Powershell:

<pre><code>dir -recurse | select-object extension -unique
</code></pre>

Thanks to <a href="http://kevin-berridge.blogspot.com/2007/11/windows-powershell.html" rel="noreferrer">http://kevin-berridge.blogspot.com/2007/11/windows-powershell.html</a>

blocks|key|1175510|text|我的awk-less，sed-less，Perl-less，Python-less+POSIX兼容的替代方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1175511|find+.+-type+f+%7C+rev+%7C+cut+-d.+-f1+%7C+rev++%7C+tr+'[:upper:]'+'[:lower:]'+%7C+sort+%7C+uniq+--count+%7C+sort+-rn|code-block|syntax|javascript|1175512|诀窍是它颠倒了这条线，并在开始时切断了延长线。|1175513|它还将扩展名转换为小写。|1175514|输出示例：|1175515|+++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|1175516|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

My awk-less, sed-less, Perl-less, Python-less POSIX-compliant alternative:

<pre><code>find . -type f | rev | cut -d. -f1 | rev | tr '[:upper:]' '[:lower:]' | sort | uniq --count | sort -rn
</code></pre>

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:

<pre><code> 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
</code></pre>

blocks|key|1390508|text|查找所有带点的内容，并且只显示后缀。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1390509|find+.+-type+f+-name+"*.*"+%7C+awk+-F.+'{print+$NF}'+%7C+sort+-u|code-block|syntax|javascript|1390510|如果您知道所有后缀都有3个字符，那么|1390511|find+.+-type+f+-name+"*.???"+%7C+awk+-F.+'{print+$NF}'+%7C+sort+-u|1390512|或with+sed显示具有一到四个字符的所有后缀。将{1,4}更改为后缀中所需的字符范围。|1390513|find+.+-type+f+%7C+sed+-n+'s/.*\.$.\{1,4\}$$/\1/p'%7C+sort+-u|1390514|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Find everythin with a dot and show only the suffix.

<pre><code>find . -type f -name "*.*" | awk -F. '{print $NF}' | sort -u
</code></pre>

if you know all suffix have 3 characters then

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

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.

<pre><code>find . -type f | sed -n 's/.*\.$.\{1,4\}$$/\1/p'| sort -u
</code></pre>

blocks|key|1390790|text|将我自己的变体添加到混合中。我认为这是最简单的，当效率不是一个大问题时，它可能会很有用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1390791|find+.+-type+f+%7C+grep+-oE+'\.(\w%2B)$'+%7C+sort+-u|code-block|syntax|javascript|1390792|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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.
<pre><code>find . -type f | grep -oE '\.(\w+)$' | sort -u
</code></pre>

blocks|key|1175223|text|在Python中，对非常大的目录使用生成器，包括空白扩展名，并获取每个扩展名出现的次数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1175224|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)|code-block|syntax|javascript|1175225|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

<pre><code>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)
</code></pre>

blocks|key|1390572|text|因为已经有了另一种使用Perl的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1390573|如果您安装了Python，您还可以(从shell)执行以下操作：|1390574|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)"|code-block|syntax|javascript|1390575|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Since there's already another solution which uses Perl:

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

<pre><code>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)"
</code></pre>

blocks|key|1175127|text|到目前为止，所有的回复都没有正确地处理带有换行符的文件名(除了ChristopheD，它是在我输入这段代码的时候进来的)。下面不是一个shell一行程序，但是可以工作，并且速度相当快。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1175128|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|code-block|syntax|javascript|1175129|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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.

<pre><code>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
</code></pre>

blocks|key|1175352|text|我认为最简单直接的方法是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1175353|for+f+in+*.*;+do+echo+"${f##*.}";+done+%7C+sort+-u|code-block|syntax|javascript|1175354|它是在ChristopheD的第三种方法上修改的。|1175355|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I think the most simple &amp; straightforward way is

<pre><code>for f in *.*; do echo "${f##*.}"; done | sort -u
</code></pre>

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

blocks|key|1390878|text|我认为这一点还没有被提及：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1390879|find+.+-type+f+-exec+sh+-c+'echo+"${0##*.}"'+{}+\;+%7C+sort+%7C+uniq+-c|code-block|syntax|javascript|1390880|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I don't think this one was mentioned yet:

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

blocks|key|1391135|text|另一种方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391136|find+.+-type+f+-name+"*.*"+-printf+"%25f\n"+%7C+while+IFS=+read+-r;+do+echo+"${REPLY##*.}";+done+%7C+sort+-u|offset|length|style|CODE|1391137|您可以删除-name+"*.*"，但这可以确保我们只处理具有某种扩展名的文件。|1391138|-printf是find的打印，而不是bash。-printf+"%25f\n"只打印文件名，去掉路径(并添加一个换行符)。|1391139|然后，我们使用字符串替换，使用${REPLY##*.}删除直到最后一个点。|1391140|注意，$REPLY只是read的内置变量，我们可以像在while+IFS=+read+-r+file中一样使用我们自己的变量，这里的变量是$file。|1391141|entityMap^0|0|0|2U|0|5|B|0|0|7|8|4|O|E|0|F|C|0|3|6|B|4|R|N|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@$D|T|E|U|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|V|8|@$D|W|E|X|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|Y|8|@$D|Z|E|10|F|G]|$D|11|E|12|F|G]|$D|13|E|14|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|15|8|@$D|16|E|17|F|G]]|9|@]|A|$]]|$1|N|3|O|5|6|7|18|8|@$D|19|E|1A|F|G]|$D|1B|E|1C|F|G]|$D|1D|E|1E|F|G]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]]|Q|$]]

Another way:
<code>find . -type f -name &quot;*.*&quot; -printf &quot;%f\n&quot; | while IFS= read -r; do echo &quot;${REPLY##*.}&quot;; done | sort -u</code>
You can drop the <code>-name &quot;*.*&quot;</code> but this ensures we are dealing only with files that do have an extension of some sort.
The <code>-printf</code> is <code>find</code>'s print, not bash. <code>-printf &quot;%f\n&quot;</code> prints only the filename, stripping the path (and adds a newline).
Then we use string substitution to remove up to the last dot using <code>${REPLY##*.}</code>.
Note that <code>$REPLY</code> is simply <code>read</code>'s inbuilt variable. We could just as use our own in the form: <code>while IFS= read -r file</code>, and here $file would be the variable.

blocks|key|1175263|text|你也可以这样做|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1175264|find+.+-type+f+-name+"*.php"+-exec+PATHTOAPP+{}+%2B|code-block|syntax|javascript|1175265|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

you could also do this

<pre><code>find . -type f -name "*.php" -exec PATHTOAPP {} +
</code></pre>

blocks|key|1391008|text|我发现它既简单又快速...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1391009|+++#+find+.+-type+f+-exec+basename+{}+\;+%7C+awk+-F"."+'{print+$NF}'+>+/tmp/outfile.txt
+++#+cat+/tmp/outfile.txt+%7C+sort+%7C+uniq+-c%7C+sort+-n+>+tmp/outfile_sorted.txt|code-block|syntax|javascript|1391010|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I've found it simple and fast...

<pre><code> # find . -type f -exec basename {} \; | awk -F"." '{print $NF}' &gt; /tmp/outfile.txt
 # cat /tmp/outfile.txt | sort | uniq -c| sort -n &gt; tmp/outfile_sorted.txt
</code></pre>

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?

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

Python

Linux

Perl

在Linux机器上，我想遍历一个文件夹层次结构，并获得其中所有不同文件扩展名的列表。从shell中实现这一点的最好方法是什么？

问如何在文件夹层次结构中找到所有不同的文件扩展名？
EN

回答 14

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问如何在文件夹层次结构中找到所有不同的文件扩展名？EN

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问如何在文件夹层次结构中找到所有不同的文件扩展名？
EN