Python:具有不同工作目录的子进程 [英] Python: subprocess with different working directory

问题描述

我在这个目录下有一个 python 脚本:

I have a python script that is under this directory:

work/project/test/a.py

在 a.py 中，我使用 subprocess.POPEN 从另一个目录启动进程，

Inside a.py, I use subprocess.POPEN to launch the process from another directory,

work/to_launch/file1.pl, file2.py, file3.py, ...

Python 代码:

subprocess.POPEN("usr/bin/perl ../to_launch/file1.pl") 

在工作/项目/下，我输入以下内容

and under work/project/, I type the following

[user@machine project]python test/a.py,

错误file2.py，'没有那个文件或目录'"

error "file2.py, 'No such file or directory'"

如何添加work/to_launch/，以便可以找到这些依赖文件file2.py?

How can I add work/to_launch/, so that these dependent files file2.py can be found?

推荐答案

您的代码不起作用，因为相对路径是相对于您的当前位置(test/a.py).

Your code does not work, because the relative path is seen relatively to your current location (one level above the test/a.py).

在 sys.path[0] 中，您有当前正在运行的脚本的路径.

In sys.path[0] you have the path of your currently running script.

使用 os.path.join(os.path.abspath(sys.path[0]), relPathToLaunch) 和 relPathToLaunch = '../to_launch/file1.pl' 获取 file1.pl 的绝对路径并使用它运行 perl.

Use os.path.join(os.path.abspath(sys.path[0]), relPathToLaunch) with relPathToLaunch = '../to_launch/file1.pl' to get the absolute path to your file1.pl and run perl with it.

编辑:如果您想从其目录启动 file1.pl 然后返回，只需记住您当前的工作目录，然后切换回来:

EDIT: if you want to launch file1.pl from its directory and then return back, just remember your current working directory and then switch back:

origWD = os.getcwd() # remember our original working directory

os.chdir(os.path.join(os.path.abspath(sys.path[0]), relPathToLaunch))
subprocess.POPEN("usr/bin/perl ./file1.pl") 
[...]

os.chdir(origWD) # get back to our original working directory

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