在Linux中打开文件夹窗口的标准方法? [英] Standard way to open a folder window in linux?

问题描述

我想在适当的文件管理器中，从跨平台(windows/mac/linux)Python应用程序中打开一个文件夹窗口.

I want to open a folder window, in the appropriate file manager, from within a cross-platform (windows/mac/linux) Python application.

在OSX上，我可以使用以下方法在查找器中打开一个窗口

On OSX, I can open a window in the finder with

os.system('open "%s"' % foldername)

，在Windows上，

and on Windows with

os.startfile(foldername)

unix/linux呢?有标准的方法吗?还是我必须对gnome/kde/etc进行特殊处理并手动运行相应的应用程序(nautilus/konqueror/etc)?

What about unix/linux? Is there a standard way to do this or do I have to special case gnome/kde/etc and manually run the appropriate application (nautilus/konqueror/etc)?

看起来可以由 freedesktop.org 人们指定的内容(类似于webbrowser的python模块，也很好！).

This looks like something that could be specified by the freedesktop.org folks (a python module, similar to webbrowser, would also be nice!).

推荐答案

os.system('xdg-open "%s"' % foldername)

xdg-open也可以用于文件/URL

xdg-open can be used for files/urls also

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