creating the most basic launch.json file is explained well in official docs.

# changing execution path

However what I really wanted to know was how to configure it to make it run debugging on a file from a specific directory path.

For example, I have a workspace structure like this.

workspace_root/
- main.py
- utils.py
- subdir1/
- t1.py

And t1.py file imports utils.py like this

import os, sys
sys.path.append(os.path.abspath('..'))

from utils import some_func1

### blahblah

In this case, t1.py is configured to run from workspace_root/subdir and not from workspace_root/, and this is not the way the default launch.json will work.

Adding cwd option fixed this problem. In this example, the following cwd option will set the debugging to run from the directory where the running file exists.

{
// Use IntelliSense to learn about possible attributes.
// Hover to view descriptions of existing attributes.
"version": "0.2.0",
"configurations": [
{
"name": "Python: Current File",
"type": "python",
"request": "launch",
"program": "${file}", "console": "integratedTerminal", "cwd": "${fileDirname}"
}
]
}

## passing arguments

If I have a python script that takes in command line arguments in order to execute them, I would also want to pass arguments when debugging such a python file. For example, if I created a python file that normally executes like this

$python test.py /path/to/config/file In this case, we can setup the vscode launch.json file like this by adding the args option { // Use IntelliSense to learn about possible attributes. // Hover to view descriptions of existing attributes. // For more information, visit: https://go.microsoft.com/fwlink/?linkid=830387 "version": "0.2.0", "configurations": [ { "name": "Python: Current File", "type": "python", "request": "launch", "program": "${file}",
"console": "integratedTerminal",
"cwd": "\${fileDirname}",
"args": ["/some/path/testconfig.yaml"]
}
]
}
Categories: python