How can I specify working directory for popen

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Is there a way to specify the running directory of command in Python's subprocess.Popen()?

For example:

Popen('c:\mytool\tool.exe', workingdir='d:\test\local')

My Python script is located in C:\programs\python

Is is possible to run C:\mytool\tool.exe in the directory D:\test\local?

How do I set the working directory for a sub-process?

popen subprocess python
2 Answers

Other way is to simply do this

cwd = os.getcwd()
os.chdir('c:\some\directory')
subprocess.Popen('tool.exe')
os.chdir(cwd)

This solution works if you want to rely on relative paths, for example, if your tool's location is c:\some\directory\tool.exe. cwd keyword argument for Popen will not let you do this. Some scripts/tools may rely on you being in the given directory while invoking them. To make this code less noisy, aka detach the logic related to changing directories from the "business logic", you can use a decorator.

def invoke_at(path: str):
    def parameterized(func):
        def wrapper(*args, **kwargs):
            cwd = os.getcwd()
            os.chdir(path)

            try:
                ret = func(*args, **kwargs)
            finally:
                os.chdir(cwd)

            return ret

        return wrapper

    return parameterized

Such decorator can be then used in a way:

@invoke_at(r'c:\some\directory')
def start_the_tool():
    subprocess.Popen('tool.exe')
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