Raf*_*ird 21 python windows asynchronous
如何在Windows上等待Python中的多个子进程,没有主动等待(轮询)?这样的东西几乎适合我:
proc1 = subprocess.Popen(['python','mytest.py'])
proc2 = subprocess.Popen(['python','mytest.py'])
proc1.wait()
print "1 finished"
proc2.wait()
print "2 finished"
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问题是,在proc2完成之前proc1,父进程仍将等待proc1.在Unix上,人们会waitpid(0)在循环中使用它来完成子进程的返回代码 - 如何在Windows上用Python实现类似的东西?
tzo*_*zot 14
这似乎有点矫枉过正,但是,这里有:
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
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基于zseil的答案,您可以通过混合使用subprocess和win32 API来完成此操作.我使用直接ctypes,因为我的Python没有安装win32api.我现在只是从MSYS中生成sleep.exe作为一个例子,但显然你可以生成任何你喜欢的进程.我使用OpenProcess()从进程'PID获取HANDLE,然后使用WaitForMultipleObjects等待任何进程完成.
import ctypes, subprocess
from random import randint
SYNCHRONIZE=0x00100000
INFINITE = -1
numprocs = 5
handles = {}
for i in xrange(numprocs):
sleeptime = randint(5,10)
p = subprocess.Popen([r"c:\msys\1.0\bin\sleep.exe", str(sleeptime)], stdin=subprocess.PIPE, stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=False)
h = ctypes.windll.kernel32.OpenProcess(SYNCHRONIZE, False, p.pid)
handles[h] = p.pid
print "Spawned Process %d" % p.pid
while len(handles) > 0:
print "Waiting for %d children..." % len(handles)
arrtype = ctypes.c_long * len(handles)
handle_array = arrtype(*handles.keys())
ret = ctypes.windll.kernel32.WaitForMultipleObjects(len(handle_array), handle_array, False, INFINITE)
h = handle_array[ret]
ctypes.windll.kernel32.CloseHandle(h)
print "Process %d done" % handles[h]
del handles[h]
print "All done!"
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