Python running as Windows Service: OSError: [WinError 6] The handle is invalid

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:

which causes the following error:

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

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Answers:

Method 1

Method 2

Answers:

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Method 1

Line 1117 in subprocess.py is:

p2cread = _winapi.GetStdHandle(_winapi.STD_INPUT_HANDLE)

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

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

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)

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)

Method 2

Add stdin=subprocess.PIPE like:

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

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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