blocks|key|1653497|text|使用Python+2.4%2B上提供的subprocess模块，而不是os.system()，这样就不必处理外壳转义。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1653498|import+subprocess,+os,+platform
if+platform.system()+==+'Darwin':+++++++#+macOS
++++subprocess.call(('open',+filepath))
elif+platform.system()+==+'Windows':++++#+Windows
++++os.startfile(filepath)
else:+++++++++++++++++++++++++++++++++++#+linux+variants
++++subprocess.call(('xdg-open',+filepath))|code-block|syntax|javascript|1653499|双括号是因为subprocess.call()想要一个序列作为它的第一个参数，所以我们在这里使用了一个元组。在带有Gnome的Linux系统上，也有一个gnome-open命令可以做同样的事情，但是xdg-open是自由桌面基础标准，可以在Linux桌面环境中工作。|1653500|entityMap^0|H|A|X|B|0|0|6|H|24|A|2R|8|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|N|$]]

Use the <code>subprocess</code> module available on Python 2.4+, not <code>os.system()</code>, so you don't have to deal with shell escaping.

<pre><code>import subprocess, os, platform
if platform.system() == 'Darwin': # macOS
 subprocess.call(('open', filepath))
elif platform.system() == 'Windows': # Windows
 os.startfile(filepath)
else: # linux variants
 subprocess.call(('xdg-open', filepath))
</code></pre>

The double parentheses are because <code>subprocess.call()</code> wants a sequence as its first argument, so we're using a tuple here. On Linux systems with Gnome there is also a <code>gnome-open</code> command that does the same thing, but <code>xdg-open</code> is the Free Desktop Foundation standard and works across Linux desktop environments.

blocks|key|1653382|text|我更喜欢：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1653383|os.startfile(path,+'open')|code-block|syntax|javascript|1653384|请注意，此模块支持文件夹和文件中包含空格的文件名，例如|1653385|A:\abc\folder+with+spaces\file+with-spaces.txt|1653386|(python+docs)不必添加'open‘(这是默认设置)。文档特别提到，这就像在Windows资源管理器中双击文件的图标一样。|offset|length|1653387|此解决方案仅适用于windows。|1653388|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/os.html#os.startfile^0|0|0|0|0|1|B|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@$M|13|N|14|1|15]]|A|$]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

I prefer:

<pre><code>os.startfile(path, 'open')
</code></pre>

Note that this module supports filenames that have spaces in their folders and files e.g. 

<pre><code>A:\abc\folder with spaces\file with-spaces.txt
</code></pre>

(<a href="http://docs.python.org/library/os.html#os.startfile" rel="noreferrer">python docs</a>) 'open' does not have to be added (it is the default). The docs specifically mention that this is like double-clicking on a file's icon in Windows Explorer.

This solution is windows only.

blocks|key|1653436|text|只是为了完整性(这不在问题中)，xdg-open将在Linux上做同样的事情。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1653437|entityMap|0|LINK|mutability|MUTABLE|url|https://portland.freedesktop.org/doc/xdg-open.html^0|G|8|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Just for completeness (it wasn't in the question), <a href="https://portland.freedesktop.org/doc/xdg-open.html" rel="noreferrer">xdg-open</a> will do the same on Linux.

blocks|key|1438844|text|import+os
import+subprocess

def+click_on_file(filename):
++++'''Open+document+with+default+application+in+Python.'''
++++try:
++++++++os.startfile(filename)
++++except+AttributeError:
++++++++subprocess.call(['open',+filename])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1438845|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
import subprocess

def click_on_file(filename):
 '''Open document with default application in Python.'''
 try:
 os.startfile(filename)
 except AttributeError:
 subprocess.call(['open', filename])
</code></pre>

blocks|key|1653569|text|Start不支持长路径名和空格。您必须将其转换为8.3兼容路径。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1653570|import+subprocess
import+win32api

filename+=+"C:\\Documents+and+Settings\\user\\Desktop\file.avi"
filename_short+=+win32api.GetShortPathName(filename)

subprocess.Popen('start+'+%2B+filename_short,+shell=True+)|code-block|syntax|javascript|1653571|该文件必须存在才能使用API调用。|1653572|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|$]]

Start does not support long path names and white spaces. You have to convert it to 8.3 compatible paths.

<pre><code>import subprocess
import win32api

filename = "C:\\Documents and Settings\\user\\Desktop\file.avi"
filename_short = win32api.GetShortPathName(filename)

subprocess.Popen('start ' + filename_short, shell=True )
</code></pre>

The file has to exist in order to work with the API call.

blocks|key|1653636|text|Windows下的os.startfile(path,+'open')是好的，因为当目录中存在空格时，os.system('start',+path_name)无法正确打开应用，当目录中存在i18n时，os.system需要将unicode更改为Windows中控制台的编解码器。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1653637|entityMap^0|9|Q|1F|T|2T|9|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

<code>os.startfile(path, 'open')</code> under Windows is good because when spaces exist in the directory, <code>os.system('start', path_name)</code> can't open the app correctly and when the i18n exist in the directory, <code>os.system</code> needs to change the unicode to the codec of the console in Windows.

blocks|key|1653599|text|我来得太晚了，但这里有一个使用windows+api的解决方案。这将始终打开关联的应用程序。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1653600|import+ctypes

shell32+=+ctypes.windll.shell32
file+=+'somedocument.doc'

shell32.ShellExecuteA(0,"open",file,0,0,5)|code-block|syntax|javascript|1653601|有很多神奇的常量。第一个零是当前程序的hwnd。可以为零。另外两个零是可选参数(参数和目录)。5+==+SW_SHOW，它指定如何执行应用程序。有关更多信息，请阅读ShellExecute+API+docs。|offset|length|1653602|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/bb762153%2528v=vs.85%2529.aspx^0|0|0|2A|L|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]]]]

I am pretty late to the lot, but here is a solution using the windows api. This always opens the associated application.

<pre><code>import ctypes

shell32 = ctypes.windll.shell32
file = 'somedocument.doc'

shell32.ShellExecuteA(0,"open",file,0,0,5)
</code></pre>

A lot of magic constants. The first zero is the hwnd of the current program. Can be zero. The other two zeros are optional parameters (parameters and directory). 5 == SW_SHOW, it specifies how to execute the app.
Read the 
<a href="http://msdn.microsoft.com/en-us/library/bb762153%28v=vs.85%29.aspx" rel="nofollow">ShellExecute API docs</a> for more info.

blocks|key|1438717|text|在mac+os上，您可以调用open|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1438718|import+os
os.open("open+myfile.txt")|code-block|syntax|javascript|1438719|这将使用TextEdit或任何设置为此文件类型的默认应用程序打开该文件。|1438720|entityMap^0|E|4|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

On mac os you can call <code>open</code>:
<pre><code>import os
os.open(&quot;open myfile.txt&quot;)
</code></pre>
This would open the file with TextEdit, or whatever app is set as default for this filetype.

blocks|key|1653546|text|如果要在Mac上指定用于打开文件的应用程序，请使用以下命令：os.system("open+-a+[app+name]+[file+name]")|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1653547|entityMap^0|U|17|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

If you want to specify the app to open the file with on Mac OS X, use this:
<code>os.system("open -a [app name] [file name]")</code>

blocks|key|1653675|text|在Windows8.1上，下面的方法可以工作，而其他给定的subprocess.call方法会失败，路径中有空格。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1653676|subprocess.call('cmd+/c+start+""+"any+file+path+with+spaces"')|code-block|syntax|javascript|1653677|通过利用这个和其他人之前的答案，这里有一个内联代码，它可以在多个平台上工作。|1653678|import+sys,+os,+subprocess
subprocess.call(('cmd+/c+start+""+"'%2B+filepath+%2B'"')+if+os.name+is+'nt'+else+('open'+if+sys.platform.startswith('darwin')+else+'xdg-open',+filepath))|1653679|entityMap^0|T|F|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

On windows 8.1, below have worked while other given ways with <code>subprocess.call</code> fails with path has spaces in it.

<pre><code>subprocess.call('cmd /c start "" "any file path with spaces"')
</code></pre>

By utilizing this and other's answers before, here's an inline code which works on multiple platforms.

<pre><code>import sys, os, subprocess
subprocess.call(('cmd /c start "" "'+ filepath +'"') if os.name is 'nt' else ('open' if sys.platform.startswith('darwin') else 'xdg-open', filepath))
</code></pre>

blocks|key|1439055|text|以下是Nick的回答，针对WSL进行了略微调整：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1439056|import+os
import+sys
import+logging
import+subprocess

def+get_platform():
++++if+sys.platform+==+'linux':
++++++++try:
++++++++++++proc_version+=+open('/proc/version').read()
++++++++++++if+'Microsoft'+in+proc_version:
++++++++++++++++return+'wsl'
++++++++except:
++++++++++++pass
++++return+sys.platform

def+open_with_default_app(filename):
++++platform+=+get_platform()
++++if+platform+==+'darwin':
++++++++subprocess.call(('open',+filename))
++++elif+platform+in+['win64',+'win32']:
++++++++os.startfile(filename.replace('/','\\'))
++++elif+platform+==+'wsl':
++++++++subprocess.call('cmd.exe+/C+start'.split()+%2B+[filename])
++++else:+++++++++++++++++++++++++++++++++++#+linux+variants
++++++++subprocess.call(('xdg-open',+filename))|code-block|syntax|javascript|1439057|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|$]]

Here is the answer from Nick, adjusted slightly for WSL:
<pre><code>import os
import sys
import logging
import subprocess

def get_platform():
 if sys.platform == 'linux':
 try:
 proc_version = open('/proc/version').read()
 if 'Microsoft' in proc_version:
 return 'wsl'
 except:
 pass
 return sys.platform

def open_with_default_app(filename):
 platform = get_platform()
 if platform == 'darwin':
 subprocess.call(('open', filename))
 elif platform in ['win64', 'win32']:
 os.startfile(filename.replace('/','\\'))
 elif platform == 'wsl':
 subprocess.call('cmd.exe /C start'.split() + [filename])
 else: # linux variants
 subprocess.call(('xdg-open', filename))
</code></pre>

blocks|key|1653738|text|我想你可能想在编辑器中打开文件。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1653739|适用于Windows|1653740|subprocess.Popen(["notepad",+filename])|code-block|syntax|javascript|1653741|对于Linux|1653742|subprocess.Popen(["text-editor",+filename])|1653743|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I think you might want to open file in editor.
For Windows
<pre><code>subprocess.Popen([&quot;notepad&quot;, filename])
</code></pre>
For Linux
<pre><code>subprocess.Popen([&quot;text-editor&quot;, filename])
</code></pre>

I need to be able to open a document using its default application in Windows and Mac OS. Basically, I want to do the same thing that happens when you double-click on the document icon in Explorer or Finder. What is the best way to do this in Python?

Open document with default OS application in Python, both in Windows and Mac OS

Windows

Python

Linux

我需要能够在Windows和Mac OS中使用默认应用程序打开文档。基本上，我想做的事情与在资源管理器或Finder中双击文档图标时发生的事情相同。在Python中做这件事的最好方法是什么？

问在Windows和Mac OS中使用Python中的默认OS应用程序打开文档
EN

回答 12

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问在Windows和Mac OS中使用Python中的默认OS应用程序打开文档EN

回答 12

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问在Windows和Mac OS中使用Python中的默认OS应用程序打开文档
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