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.
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]
.
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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