Python：具有不同工作目录的子进程
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Python: subprocess with different working directory
提问by pepero
I have a python script that is under this directory:
我在这个目录下有一个 python 脚本：
work/project/test/a.py
Inside a.py, I use subprocess.POPENto launch the process from another directory,
在里面a.py，我用来subprocess.POPEN从另一个目录启动进程，
work/to_launch/file1.pl, file2.py, file3.py, ...
Python Code:
蟒蛇代码：
subprocess.POPEN("usr/bin/perl ../to_launch/file1.pl")
and under work/project/, I type the following
在工作/项目/下，我输入以下内容
[user@machine project]python test/a.py,
error "file2.py, 'No such file or directory'"
错误“file2.py，'没有这样的文件或目录'”
How can I add work/to_launch/, so that these dependent files file2.pycan be found?
如何添加work/to_launch/，以便file2.py可以找到这些依赖文件？
回答by eumiro
Your code does not work, because the relative path is seen relatively to your current location (one level above the test/a.py).
您的代码不起作用，因为相对路径是相对于您的当前位置（在 上的一级test/a.py）。
In sys.path[0]you have the path of your currently running script.
在sys.path[0]你有你的当前运行的脚本的路径。
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.pland run perlwith it.
使用os.path.join(os.path.abspath(sys.path[0]), relPathToLaunch)withrelPathToLaunch = '../to_launch/file1.pl'获取你的绝对路径file1.pl并运行perl它。
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:
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：
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
回答by anijhaw
You could use this code to set the current directory:
您可以使用此代码设置当前目录：
import os
os.chdir("/path/to/your/files")
回答by Adam Byrtek
Use paths relative to the script, not the current working directory
使用相对于脚本的路径，而不是当前工作目录
os.path.join(os.path.dirname(__file__), '../../to_launch/file1.pl)
See also my answer to Python: get path to file in sister directory?
另请参阅我对Python 的回答：获取姊妹目录中的文件路径？
