如何在 Visual Studio Code 的 launch.json 中调试 Python 模块

Question

My question may seem simple but, I have a module that I launch in a terminal like this:

python -m my_module.my_file

How do I debug this in Visual Studio Code?

I have this in my launch.json ( documentation )

"type": "python",
"request": "launch",
"pythonPath": "D:\\ProgramData\\Anaconda3\\envs\\simulec\\python.exe",
"args": [
   "-m",
   "my_module.my_file",
]

If I don't set the program option or if I set it to "" I get "File does not exist" Error.

How can I fix this?

Answer 1

Actually, there is a very simple option to do this I found by accident while trying to edit the launch.json file.

"type": "python",
"request": "launch",
"pythonPath": "D:\\ProgramData\\Anaconda3\\envs\\simulec\\python.exe",
"module": "my_module.my_file",

Simply specify the module in the module key "module": "my_module.my_file"

The -m is not useful any more.

Answer 2

In a terminal like this: python -m xyz abc.z3

(Please make sure you are opening "the root folder of your project").

{
    // 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": "module",
            "type": "python",
            "request": "launch",
            "module": "xyz",
            "args": [
                "abc.z3"
            ]
        }
    ],
}

Answer 3

The answer from @dzada was a bit confusing for me so I tried to rephrase it and add more clarification.

To debug a module in file called script_file.py that exists in a package called packagex with structure like the following:

Project_Folder
   packagex
      script_file.py

Your configuration in the launch.json file should looks like the following

    "configurations": [
    {
        "name": "Python: any name like script_file",
        "type": "python",
        "request": "launch",
        "module": "packagex.script_file"
    }
]

Answer 4

To add slightly to dzada's answer , which helped me a lot, a Visual Studio Code variable can be used to make this general for debugging any file that is in your module.

{
    "version": "0.2.0",
    "configurations": [
        {
            "name": "Python: Module",
            "type": "python",
            "request": "launch",
            "module": "my_module.${fileBasenameNoExtension}"
        }
    ]
}

Which is probably what you want to do.

Answer 5

After some searching, I found that the config can be generalized for any module by the use of the Command Variable extension . Assuming that the your workspace folder is the one containing the package, the following configuration would work for any module:

{
    "name": "Python: Module",
    "type": "python",
    "request": "launch",
    "justMyCode": true,
    "module": "${command:extension.commandvariable.file.relativeFileDotsNoExtension}",
    "cwd":"${workspaceFolder}"
}

Answer 6

{
    "version": "0.2.0",
    "configurations": [
        {
            "name": "module",
            "type": "pythonExperimental",
            "request": "launch",
            "module": "my_package.my_module.${fileBasenameNoExtension}",
        },
    ]
}