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I don't have a clue on where to look for a solution to this. Basically i want to make a file that can be "opened with" a python script. The directory of the opened file will be stored as a variable in the script. Here's what I mean by "Open With":

Sorry if I'm not too good at describing this.

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  • Do you want to open .phktest file with c:\some\path\phktestrunner.py or you want to open it with only python? Commented Aug 11, 2020 at 6:33
  • i want to open .phktest with the Python FILE, not Python itself. It would be nice if I could store the full path of the .phktest file being opened as a variable. i.e. C:\!PythonCode2\runfileswith\test.phktest Commented Aug 11, 2020 at 6:37
  • By Python FILE you meant a file with .py extension. Commented Aug 11, 2020 at 6:45

2 Answers 2

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There might be a more direct way, but one approach could be to write a simple batch file that invokes the Python interpreter with your script:

main.bat

@echo off
python main.py %1

Add absolute or relative paths if necessary.

Within main.py, the name of the file should then be available as sys.argv[1].

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This does technically work, but I'm not able to see any Python errors, because the console closes immediately. I've tried adding /K to the .bat file but still nothing
Add pause at the end of the batch file, after the python command line.
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If I'm not misunderstanding, what you want is your argv[1] be 'C:!PythonCode2\runfileswith\test.phktest'.

I did some test on my Windows 10, simply drag the .phktest file onto the main.py will let sys.argv[1] be the full path of that file!

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