blocks|key|692124|text|假设m是一个字符串，您可以使用endswith|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|692125|if+m.endswith('.mp3'):
...
elif+m.endswith('.flac'):
...|code-block|syntax|javascript|692126|不区分大小写，并消除潜在的较大的else-if链：|692127|m.lower().endswith(('.png',+'.jpg',+'.jpeg'))|692128|entityMap^0|2|1|F|8|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

Assuming <code>m</code> is a string, you can use <code>endswith</code>:

<pre><code>if m.endswith('.mp3'):
...
elif m.endswith('.flac'):
...
</code></pre>

To be case-insensitive, and to eliminate a potentially large else-if chain:

<pre><code>m.lower().endswith(('.png', '.jpg', '.jpeg'))
</code></pre>

blocks|key|908366|text|从Python3.4开始使用pathlib。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|908367|from+pathlib+import+Path
Path('my_file.mp3').suffix+==+'.mp3'|code-block|syntax|javascript|908368|entityMap^0|E|7|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|$]]

Use <code>pathlib</code> From Python3.4 onwards.

<pre><code>from pathlib import Path
Path('my_file.mp3').suffix == '.mp3'
</code></pre>

blocks|key|908199|text|看看模块fnmatch。它会做你想做的事。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|908200|import+fnmatch
import+os

for+file+in+os.listdir('.'):
++++if+fnmatch.fnmatch(file,+'*.txt'):
++++++++print+file|code-block|syntax|javascript|908201|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|$]]

Look at module fnmatch. That will do what you're trying to do.

<pre><code>import fnmatch
import os

for file in os.listdir('.'):
 if fnmatch.fnmatch(file, '*.txt'):
 print file
</code></pre>

blocks|key|692138|text|或者：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|692139|from+glob+import+glob
...
for+files+in+glob('path/*.mp3'):+
++do+something
for+files+in+glob('path/*.flac'):+
++do+something+else|code-block|syntax|javascript|692140|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|$]]

or perhaps: 

<pre><code>from glob import glob
...
for files in glob('path/*.mp3'): 
 do something
for files in glob('path/*.flac'): 
 do something else
</code></pre>

blocks|key|692289|text|一种简单的方法可能是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|692290|import+os

if+os.path.splitext(file)[1]+==+".mp3":
++++#+do+something|code-block|syntax|javascript|692291|os.path.splitext(file)将返回一个包含两个值的元组(不带扩展名的文件名%2B只带扩展名)。因此，第二个索引(1)将只给出扩展。最酷的是，如果需要的话，用这种方式你也可以很容易地访问文件名！|offset|length|style|CODE|692292|entityMap^0|0|0|0|M|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

one easy way could be:

<pre><code>import os

if os.path.splitext(file)[1] == ".mp3":
 # do something
</code></pre>

<code>os.path.splitext(file)</code> will return a tuple with two values (the filename without extension + just the extension). The second index ([1]) will therefor give you just the extension. The cool thing is, that this way you can also access the filename pretty easily, if needed!

blocks|key|908227|text|import+os
source+=+['test_sound.flac','ts.mp3']

for+files+in+source:
+++fileName,fileExtension+=+os.path.splitext(files)
+++print+fileExtension+++#+Print+File+Extensions
+++print+fileName+++#+It+print+file+name|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|908228|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>import os
source = ['test_sound.flac','ts.mp3']

for files in source:
 fileName,fileExtension = os.path.splitext(files)
 print fileExtension # Print File Extensions
 print fileName # It print file name
</code></pre>

blocks|key|692209|text|#!/usr/bin/python

import+shutil,+os

source+=+['test_sound.flac','ts.mp3']

for+files+in+source:
++fileName,fileExtension+=+os.path.splitext(files)

++if+fileExtension==".flac"+:
++++print+'This+file+is+flac+file+%25s'+%25files
++elif++fileExtension==".mp3":
++++print+'This+file+is+mp3+file+%25s'+%25files
++else:
++++print+'Format+is+not+valid'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|692210|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>#!/usr/bin/python

import shutil, os

source = ['test_sound.flac','ts.mp3']

for files in source:
 fileName,fileExtension = os.path.splitext(files)

 if fileExtension==".flac" :
 print 'This file is flac file %s' %files
 elif fileExtension==".mp3":
 print 'This file is mp3 file %s' %files
 else:
 print 'Format is not valid'
</code></pre>

blocks|key|692319|text|if+(file.split(".")[1]+==+"mp3"):
++++print+"its+mp3"
elif+(file.split(".")[1]+==+"flac"):
++++print+"its+flac"
else:
++++print+"not+compat"|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|692320|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>if (file.split(".")[1] == "mp3"):
 print "its mp3"
elif (file.split(".")[1] == "flac"):
 print "its flac"
else:
 print "not compat"
</code></pre>

blocks|key|908481|text|如果您的文件已上传，则|type|unstyled|depth|inlineStyleRanges|entityRanges|data|908482|import+os


file=+request.FILES['your_file_name']++++++++++#Your+input+file_name+for+your_file_name
ext+=+os.path.splitext(file.name)[-1].lower()


if+ext=='.mp3':
++++#do+something

elif+ext=='.xls'+or+'.xlsx'+or+'.csv':
++++#do+something

else:
++++#The+uploaded+file+is+not+the+required+format|code-block|syntax|javascript|908483|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|$]]

If your file is uploaded then
<pre><code>import os


file= request.FILES['your_file_name'] #Your input file_name for your_file_name
ext = os.path.splitext(file.name)[-1].lower()


if ext=='.mp3':
 #do something

elif ext=='.xls' or '.xlsx' or '.csv':
 #do something

else:
 #The uploaded file is not the required format
</code></pre>

blocks|key|908394|text|file='test.xlsx'
if+file.endswith('.csv'):
++++print('file+is+CSV')
elif+file.endswith('.xlsx'):
++++print('file+is+excel')
else:
++++print('none+of+them')|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|908395|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>file='test.xlsx'
if file.endswith('.csv'):
 print('file is CSV')
elif file.endswith('.xlsx'):
 print('file is excel')
else:
 print('none of them')
</code></pre>

I'm working on a certain program where I need to do different things depending on the extension of the file. Could I just use this?

<pre><code>if m == *.mp3
 ...
elif m == *.flac
 ...
</code></pre>

How can I check the extension of a file?

我正在做一个特定的程序，我需要根据文件的扩展名做不同的事情。我能用这个吗？if m == *.mp3   ...elif m == *.flac   ...

问如何检查文件的扩展名？
EN

回答 10

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何检查文件的扩展名？EN

回答 10

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何检查文件的扩展名？
EN