vscode python launch 'current file' as module

Question

I am trying to find a way to give current file as argument for module. I couldn't find it anywhere so I'm trying stackoverflow

{
    "name": "Python: Synapse",
    "type": "python",
    "request": "launch",
    "module": "projFolder.targetModule",
    "console": "integratedTerminal"
}

Manually specifying the module argument such as above example, works. But thinking of doing that for every each file wanted me to automate it. I've tried ${file}, but it gives the file path not the module. So it didn't work. How can I launch current current file as module?

Is this what you are looking for: stackoverflow.com/questions/57392698/… — rioV8
– rioV8, Commented Aug 12, 2019 at 7:49
Possible duplicate of Problem with debugging python modules in VSCode — Inyoung Kim 김인영
– Inyoung Kim 김인영, Commented Aug 13, 2019 at 6:42

Jackiexiao · Accepted Answer · 2020-12-14 16:11:04Z

11

ps: answer from https://stackoverflow.com/a/57416114/6088796, but I modify a litter

add something like this to your launch.json, but if you want to run multi-module(in a different place, the only way you can do is using multi configuration, or use a plugin(see below)

    "configurations": [
        {
            "name": "Python: Module",
            "type": "python",
            "request": "launch",
            "module": "yourmodule.${fileBasenameNoExtension}",
            "stopOnEntry": true
        },

You can use the extension Command Variable to get this relative directory with dot separator.

{
  "version": "0.2.0",
  "configurations": [
    {
      "name": "Python: Module CmdVar",
      "type": "python",
      "request": "launch",
      "console": "integratedTerminal",
      "module": "${command:extension.commandvariable.file.relativeDirDots}.${fileBasenameNoExtension}",
    }
  ]
}

answered Dec 14, 2020 at 16:11

Jackiexiao

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1 Comment

vdi Over a year ago

there is a combined command: "module": "${command:extension.commandvariable.file.relativeFileDotsNoExtension}",

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