如何制作一个python文件副本，然后执行另一个副本一次？

如何解决如何制作一个python文件副本，然后执行另一个副本一次？

我在学校项目中使用 python，我想知道如何用简单的这种效果创建一个简单的病毒

我试过这个： How do I use shutil to make a python file copy itself after doing a calculation?

还有这个： https://stackoverflow.com/questions/1186789/what-is-the-best-way-to-call-a-script-from-another-script

然后我不知道如何运行第二个副本

代码示例（我想运行一次副本）

import os
import shutil

##############

print("hello world")

shutil.copy(__file__,"copy.py") # copies the file
exec(open("copy.py").read())# Executes the copied file if it is in the same directory as the other file

解决方法

您可以使用 execfile(File Path)

运行第二个文件

所以你得到了类似的东西

import os
import shutil
import fileinput

##############

print("hello world")

shutil.copy(__file__,"copy.py") # Copies the file

file = [Path of the copied File]    

for line in fileinput.FileInput(file,inplace=1):
    if "execfile('copy.py')" in line:
        line=line.rstrip()
        line=line.replace(line,"")
    print (line,end="")
# This could get a problem with large files but should work properly with small files

execfile('copy.py') # Executes the copied file if it is in the same directory as the other file
# Else give it the directory where the file is stored (with / instead of \)

,

所以我明白了（我也在 reddit 上发过）

链接：https://www.reddit.com/r/learnpython/comments/lxupy9/how_to_make_a_python_file_copy_itself_then/

import shutil
import subprocess
import os
import random

print("hello world")
new_file_name = str(random.randint(1,10000)) + ".py"
shutil.copy(__file__,new_file_name)

subprocess.call((new_file_name),shell=True)