Python作为Windows服务运行:OSError:句柄无效

我有一个Python脚本,作为Windows服务运行。 该脚本用另一个进程分叉:

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

这会导致以下错误:

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

subprocess.py中的第1117行是:

 p2cread = _winapi.GetStdHandle(_winapi.STD_INPUT_HANDLE) 

这使我怀疑服务流程没有与他们相关的STDIN(TBC)

这个麻烦的代码可以通过提供一个文件或null设备作为stdin参数来避免popen

Python 3.3,3.4和3.5中 ,您只需传递stdin=subprocess.DEVNULL 。 例如

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

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

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

添加stdin=subprocess.PIPE如:

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