why do I have to add a higher folder when opening a file in python when debugging?

Asked Jan 02 '23 at 17:42

Active Jan 02 '23 at 18:41

Viewed 42 times

I have a directory tree similar to this:

my_application
    my_component
        process_config.py
        process_config.yaml
    runner.py

I launch python runner.py and it calls a class inside process_config.py that reads the yaml file like this:

with open(
   os.path.join("my_component", "process_config.yaml"), "r"
) as proces_config_defaults_source:

And it works fine when y run python runner.py, but it can't locate the file when I run the file from the debugger. I have to use this option for it to work:

with open(
    os.path.join("my_application", "my_component", "process_config.yaml"), "r"
) as proces_config_defaults_source:

How can I make the debugger to work the same as a python run?

edited Jan 02 '23 at 18:41

asked Jan 02 '23 at 17:42

HuLu ViCa

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I don't think the debugger is your immediate problem here. If your file exists directly next to `runner.py`, you should use a relative path, anyway, not refer to the parent folders – OneCricketeer Jan 02 '23 at 17:53
@OneCricketeer you're right, mi description of the issue was wrong. Sorry about it – HuLu ViCa Jan 02 '23 at 18:41
2

Could you please attach your `launch.json` to the issue, if you use `"program": "${file}",`, will it be solved? – MingJie-MSFT Jan 03 '23 at 01:50
Related - https://stackoverflow.com/q/61586037/2308683 – OneCricketeer Jan 03 '23 at 14:25

why do I have to add a higher folder when opening a file in python when debugging?

0 Answers0