Windows上的Python-如何等待多个子进程？

Question

如何在Windows上的Python中等待多个子进程，而无需主动等待（轮询）？这样的事情 几乎 对我有用：

proc1 = subprocess.Popen(['python','mytest.py'])
proc2 = subprocess.Popen(['python','mytest.py'])    
proc1.wait()
print "1 finished"
proc2.wait()
print "2 finished"

问题是，当proc2之前完成时proc1，父进程仍将等待proc1。在Unix上，人们会waitpid(0)在循环中使用它们来获取子进程的返回代码，这些代码在完成时如何在Windows上的Python中实现类似功能？

Answer 1

可能看起来有些矫kill过正，但是，在这里：

import Queue, thread, subprocess

results= Queue.Queue()
def process_waiter(popen, description, que):
    try: popen.wait()
    finally: que.put( (description, popen.returncode) )
process_count= 0

proc1= subprocess.Popen( ['python', 'mytest.py'] )
thread.start_new_thread(process_waiter,
    (proc1, "1 finished", results))
process_count+= 1

proc2= subprocess.Popen( ['python', 'mytest.py'] )
thread.start_new_thread(process_waiter,
    (proc2, "2 finished", results))
process_count+= 1

# etc

while process_count > 0:
    description, rc= results.get()
    print "job", description, "ended with rc =", rc
    process_count-= 1