Python popen sysargv

Question

我想用popen打开一个脚本，其中包含一个sysargv参数，如下所示：

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)

我得到了一个错误：

无法打开文件'/home/Network_Monitor_Device/Scripts/Traceroute.py 192.168.76.1'：Errno 2没有这样的文件或目录

Answer 1

它在等待一份名单。尝试：

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