Python 打开系统参数

Question

我想用 popen 打开一个脚本，带有这样的 sysargv 参数：

import subprocess

Script = '/home/Network_Monitor_Device/Scripts/Traceroute.py 192.168.76.1'

p = subprocess.Popen(['python','-u',Script], stdout = subprocess.PIPE, stderr=subprocess.STDOUT)

out = p.stdout.readline()

print out

Traceroute.py

import os
import sys
import subprocess

subprocess.check_output("traceroute " + str(sys.argv[1]), shell=True)

我收到此错误：

python: can't open file '/home/Network_Monitor_Device/Scripts/Traceroute.py 192.168.76.1': [Errno 2] No such file or directory

Answer 1

它需要一个列表。尝试：

p = subprocess.Popen(['python','-u', '/home/Network_Monitor_Device/Scripts/Traceroute.py', '192.168.76.1'], stdout = subprocess.PIPE, stderr=subprocess.STDOUT)

Python 打开系统参数

Python popen sysargv

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