我如何在较低的目录中运行另一个 Python 文件

Question

So I'm creating a program that reads itself, creates itself in every lower folder currently available and then runs each of those, in turn, copying itself lower and lower until it has all available directories below it with this program in it

 from glob import glob
 import os
 paths = glob('*/')
 file = open("Egg.py","r")
 code = ""
 for i in file:
     code = code + i
 file.close()
 for i in paths:
     name = "Egg.py"
     name = i + name
     file = open(name,"w+")
     file.write(code)
     file.close()
     #This is where i want to execute
     # Which has the name of the file location
     # followed by Egg.py
 print("Done")

this is the code I have so far, the file is called 'Egg.py' and creates the same script in the folders beneath it no problem. The issue I have is I cant then run these files as I create them to put the code in the files beneath them. Does anybody know a quick and easy way to do this? I would prefer if it was less code than more but any help at all is appreciated thanks. for reference I have the code file next to another file like this: Folder Egg.py and in the folder I have 3 other folders FolderA FolderB FolderC etc thanks again

Answer 1

If I understand correctly, there is no need for recursion here. You can walk down the current path using the following method:

import os
walk = os.walk('.')

Then, you iterate over this walk, copying your file in every directory:

# Read the contents of your current script
with open("Egg.py","r") as file:
    code = file.read()

# Explore your folders & subfolders iteratively
for root, foldernames, filenames in walk:
    with open(os.path.join(root, "Egg.py"), "w") as file:
        file.write(code)