在 python 脚本中启动 shell 命令，等待终止并返回脚本

2022-01-18 00:00:00 python subprocess popen os.execl

问题描述

我有一个 python 脚本，它必须为目录中的每个文件启动一个 shell 命令:

I've a python script that has to launch a shell command for every file in a dir:

import os

files = os.listdir(".")
for f in files:
    os.execlp("myscript", "myscript", f)

这对于第一个文件很好，但是在myscript"命令结束后，执行停止并且不会返回到 python 脚本.

This works fine for the first file, but after the "myscript" command has ended, the execution stops and does not come back to the python script.

我该怎么办?我必须在 调用 os.execlp() 之前 fork() 吗?

How can I do? Do I have to fork() before calling os.execlp()?

解决方案

subprocess:subprocess 模块允许您生成新进程，连接到他们的输入/输出/错误管道，并获取它们的返回码.

subprocess: The subprocess module allows you to spawn new processes, connect to their input/output/error pipes, and obtain their return codes.

http://docs.python.org/library/subprocess.html

用法:

import subprocess
process = subprocess.Popen(command, shell=True, stdout=subprocess.PIPE)
process.wait()
print process.returncode

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