python multiprocessing.Process执行错误的目标(与py2exe打包)

问题描述

我有问题.

i在Windows7(64位)上使用python(2.7.7，32位)和py2exe(0.6.9).

i using python(2.7.7, 32bit) and py2exe(0.6.9) on Windows7(64bit).

我的应用程序结构如下:

my application structure such as the following:

from multiprocessing import Process

def child():
  print "child"

def main():
  print "main"
  p = Process(target=child)
  p.start()
  p.join()

if __name__ == "__main__":
  main()

(1)打包前的结果:

main
child

(2)打包后的结果:

main
main
main
...(forever)

我想在包装后得到(1).

i want to get (1) after packaging.

请告诉我包装后如何获得(1).

please tell me how to get (1) after packaging.

爱.

推荐答案

如注释中所述，将Python脚本打包到Windows可执行文件中时，需要调用multiprocessing.freeze_support().此调用应在if __name__ == '__main__':之后，然后才实际调用main().

As mentioned in the comments, you need a call to multiprocessing.freeze_support() when packaging a Python script into an executable for use on Windows. This call should come just after if __name__ == '__main__': before actually calling main().

链接到multiprocessing.freeze文档

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