Python 作为 Windows 服务运行:OSError: [WinError 6] 句柄无效

Python running as Windows Service: OSError: [WinError 6] The handle is invalid(Python 作为 Windows 服务运行:OSError: [WinError 6] 句柄无效)

问题描述

我有一个 Python 脚本，它作为 Windows 服务运行.该脚本通过以下方式分叉另一个进程:

I have a Python script, which is running as a Windows Service. The script forks another process with:

with subprocess.Popen( args=[self.exec_path], stdout=subprocess.PIPE, stderr=subprocess.STDOUT) as proc:

导致以下错误:

OSError: [WinError 6] The handle is invalid
   File "C:Program Files (x86)Python35-32libsubprocess.py", line 911, in __init__
   File "C:Program Files (x86)Python35-32libsubprocess.py", line 1117, in _get_handles

推荐答案

subprocess.py 中的第 1117 行是:

Line 1117 in subprocess.py is:

p2cread = _winapi.GetStdHandle(_winapi.STD_INPUT_HANDLE)

这让我怀疑服务进程没有与之关联的 STDIN (TBC)

which made me suspect that service processes do not have a STDIN associated with them (TBC)

可以通过将文件或空设备作为标准输入参数提供给 popen 来避免这种麻烦的代码.

This troublesome code can be avoided by supplying a file or null device as the stdin argument to popen.

在 Python 3.x 中，您可以简单地传递 stdin=subprocess.DEVNULL.例如

In Python 3.x, you can simply pass stdin=subprocess.DEVNULL. E.g.

subprocess.Popen( args=[self.exec_path], stdout=subprocess.PIPE, stderr=subprocess.STDOUT, stdin=subprocess.DEVNULL)

在 Python 2.x 中，您需要将文件处理程序设为 null，然后将其传递给 popen:

In Python 2.x, you need to get a filehandler to null, then pass that to popen:

devnull = open(os.devnull, 'wb')
subprocess.Popen( args=[self.exec_path], stdout=subprocess.PIPE, stderr=subprocess.STDOUT, stdin=devnull)

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