Call to operating system to open url?

2019-01-13 14:53发布

What can I use to call the OS to open a URL in whatever browser the user has as default? Not worried about cross-OS compatibility; if it works in linux thats enough for me!

5条回答
Lonely孤独者°
2楼-- · 2019-01-13 14:55

Here is how to open the user's default browser with a given url:

import webbrowser

webbrowser.open(url[, new=0[, autoraise=True]])

Here is the documentation about this functionality. It's part of Python's stdlibs:

http://docs.python.org/library/webbrowser.html

I have tested this successfully on Linux, Ubuntu 10.10.

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对你真心纯属浪费
3楼-- · 2019-01-13 14:58

Have a look at the webbrowser module.

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Root(大扎)
4楼-- · 2019-01-13 14:59

Then how about mixing codes of @kobrien and @bobince up:

import subprocess
import webbrowser
import sys

url = 'http://test.com'
if sys.platform == 'darwin':    # in case of OS X
    subprocess.Popen(['open', url])
else:
    webbrowser.open_new_tab(url)
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戒情不戒烟
5楼-- · 2019-01-13 15:08

Personally I really wouldn't use the webbrowser module.

It's a complicated mess of sniffing for particular browsers, which will won't find the user's default browser if they have more than one installed, and won't find a browser if it doesn't know the name of it (eg Chrome).

Better on Windows is simply to use the os.startfile function, which also works on a URL. On OS X, you can use the open system command. On Linux there's xdg-open, a freedesktop.org standard command supported by GNOME, KDE and XFCE.

if sys.platform=='win32':
    os.startfile(url)
elif sys.platform=='darwin':
    subprocess.Popen(['open', url])
else:
    try:
        subprocess.Popen(['xdg-open', url])
    except OSError:
        print 'Please open a browser on: '+url

This will give a better user experience on mainstream platforms. You could fall back to webbrowser on other platforms, perhaps. Though most likely if you're on an obscure/unusual/embedded OS where none of the above work, chances are webbrowser will fail too.

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ゆ 、 Hurt°
6楼-- · 2019-01-13 15:10

You can use the webbrowser module.

webbrowser.open(url)
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