如何在 vscode 中为 Python 执行添加命令行选项？

Question

It was my understanding that to add a command line option (in my case I want the -i option) to the python execution in VScode, I was to alter the variable "python.terminal.launchArgs"

As this is curiously not working, I am going to describe the steps I took, so that perhaps some more knowledgeable person can spot my mistake.

1. I went to file > preferences > settings

2. then I searched for python

3. The first mention I found to settings.json was in Python > Analysis:Disabled. I clicked "Edit in settings.json", hoping to open the config file

4. I set the config file as follows:

    { "window.zoomLevel": 3, "python.pythonPath": "/usr/bin/python3", "python.terminal.launchArgs": ["-i"] }

5. I went back to my code and pressed F5 to run it. It ran on the terminal that comes with vscode. The command line was as follows

   lucas@mongolia:~/aulas_ed$ cd /home/lucas/aulas_ed ; env "PYTHONIOENCODING=UTF-8" "PYTHONUNBUFFERED=1" /usr/bin/python3 /home/lucas/.vscode/extensions/ms-python.python-2018.12.1/pythonFiles/ptvsd_launcher.py --default --client --host localhost --port 34661 /home/lucas/aulas_ed/teste.py

6. As you can see, that did not include "-i" .

7. I tried altering other variables in settings.json , in the same window. The text size works (text size changes immediately) and the python path variable works (if I type a nonsensical path, F5 stops working and VScode starts complaing about an undefined path -- just to say that the settings file seems to work properly for other ends.

Answer 1

Found a way... But for sure it's not THE way...

.vscode/settings.json

{
    "python.terminal.launchArgs": [ 
        "${workspaceFolder}/main.py",
        "--port 8010"
    ]
}

pythonfile (main.py)

import sys
port = int(sys.argv[-2].split(' ')[-1]) if int(sys.argv[-2].split(' ')[-1]) else 8001

# >>>  * Running on http://10.100.100.75:8010/ (Press CTRL+C to quit)