从python中的特定程序打开文件 [英] Open a file from a specific program from python

问题描述

我想做一件非常简单的事情，但我很迷茫.

I would like to do a very simple thing but I am quite lost.

我正在使用一个名为Blender的程序，我想在python中编写一个脚本以打开一个.blend文件，但要使用blender.app，该文件与blender文件位于同一文件夹中，而不是与blender.app位于应用程序"中. (使用Macosx)

I am using a program called Blender and I want to write a script in python which open a .blend file but using the blender.app which is located in the same folder with the blend file, not with the blender.app which is located in Applications. (using Macosx)

所以我在想这应该做...但是相反，它打开了搅拌机两次...

So I was thinking that this should do the job...but instead it opens blender twice...

import os

path = os.getcwd()
print(path)
os.system("cd path/")
os.system("open blender.app Import_mhx.blend")

我也尝试过这个

import os

path = os.getcwd()
print(path)
os.system("cd path/")
os.system("open Import_mhx.blend")

但是不幸的是，它使用位于应用程序中的默认blender.app打开了.blend文件.

but unfortunately it opens the .blend file with the default blender.app which is located in Applications...

有什么主意吗?

推荐答案

由于system命令在子shell中执行，并且chdir仅对该子shell有效，因此此方法不起作用.将命令替换为

This cannot work since the system command gets executed in a subshell, and the chdir is only valid for that subshell. Replace the command by

os.system("open -a path/blender.app Import_mhx.blend")

或(更好)

subprocess.check_call(["open", "-a", os.path.join(path, "blender.app"),
                       "Import_mhx.blend"])

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