如何使子进程每5秒挂起外部程序的执行？

Question

给出一个外部程序，在本例中是python target.py

target.py

import time, itertools
A = itertools.count()
while True:
    time.sleep(.1)
    print A.next()

我正在寻找一种运行该命令的方法，我们可以假设除了启动和停止之外，我无法控制该命令5秒。此时，我想挂起执行(类似于我的目标平台linux上的control-Z )，运行一些内部代码，然后继续执行子进程。到目前为止我有

reader.py

import subprocess, signal, time

cmd = "python target.py"
P = subprocess.Popen(cmd,shell=True)

while True:
   time.sleep(5)
   signal.pause(P)  # Not the correct way to suspend P
   print "doing something"
   signal.wakeup(P) # What is called here?

Answer 1

由于您是在Linux上，所以可以使用以下reader.py：

import subprocess, signal, time, os

cmd = "python target.py"
P = subprocess.Popen(cmd,shell=True)

while True:
   time.sleep(5)
   os.kill(P.pid, signal.SIGSTOP)
   print "doing something"
   os.kill(P.pid, signal.SIGCONT)

Answer 2

您也可以使用psutil并避免可怕的os.kill：

import psutil, time, subprocess

cmd = "python target.py"
P = subprocess.Popen(cmd,shell=True)
psProcess = psutil.Process(pid=P.pid)

while True:
    time.sleep(5)
    psProcess.suspend()
    print 'I am proactively leveraging my synergies!'
    psProcess.resume()

Answer 3

我们可以使用子进程的send_signal函数。

import subprocess
import signal
import time

p = subprocess.Popen(['mpg123', '-C', 'music.mp3'])
time.sleep(3)
print('stop')
p.send_signal(signal.SIGSTOP)
time.sleep(3)
print('continue')
p.send_signal(signal.SIGCONT)
time.sleep(3)
p.terminate()