的Python POPEN sysargv

Question

我要开POPEN一个脚本，用sysargv的说法是这样的：

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) 

我得到这个错误：

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

它期待一个列表。尝试：

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