Python:如何以非阻塞方式读取子进程的标准输出 [英] Python: How to read stdout of subprocess in a nonblocking way

问题描述

我正在尝试制作一个简单的 python 脚本来启动一个子进程并监视其标准输出.以下是代码片段:

I am trying to make a simple python script that starts a subprocess and monitors its standard output. Here is a snippet from the code:

process = subprocess.Popen([path_to_exe, os.path.join(temp_dir,temp_file)], stdout=subprocess.PIPE)
while True:   
    output=process.stdout.readline()
    print "test"

问题在于脚本在 output=process.stdout.readline() 上挂起，并且 print "test" 行仅在子进程终止后执行.

The problem is that the script hangs on output=process.stdout.readline() and that the line print "test" only executes after the subprocess is terminated.

有没有办法读取标准输出并打印它而不必等待子进程终止?

Is there a way to read standard output and print it without having to wait for the subprocess to terminate?

我要启动的子进程是一个 Windows 二进制文件，我没有它的源代码.

The subprocess which I am starting is a Windows binary for which I do not have the source code.

我发现了几个类似的问题，但答案仅适用于 Linux 或如果我有我要启动的 suprocess 的来源.

I have found several similar questions, but the answers are only applicable on Linux or in case I have the source of the suprocess I am starting.

推荐答案

勾选select 模块

import subprocess
import select
import time
    
x=subprocess.Popen(['/bin/bash','-c',"while true; do sleep 5; echo yes; done"],stdout=subprocess.PIPE)
    
y=select.poll()
y.register(x.stdout,select.POLLIN)

while True:
  if y.poll(1):
     print x.stdout.readline()
  else:
     print "nothing here"
     time.sleep(1)

---

非 posix 系统的线程解决方案:

---

Threaded Solution for non posix systems:

import subprocess
from threading import Thread 
import time
 
linebuffer=[]
x=subprocess.Popen(['/bin/bash','-c',"while true; do sleep 5; echo yes; done"],stdout=subprocess.PIPE)

def reader(f,buffer):
   while True:
     line=f.readline()
     if line:
        buffer.append(line)
     else:
        break

t=Thread(target=reader,args=(x.stdout,linebuffer))
t.daemon=True
t.start()

while True:
  if linebuffer:
     print linebuffer.pop(0)
  else:
     print "nothing here"
     time.sleep(1)

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