Question 134289 in Visual Studio Code Question posted in
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Visual Studio Code – Launch a program in vscode only if a certain condition is satisfied, e.g. the filename matches. (using launch.json)

I am trying to write a launch.json config to run a python program called myProg. By design, myProg needs to be run from a folder containing a file named myInput.py; My workflow is that I open myInput.py, put some breakpoints there, and launch myProg through the following launch configuration:

{
  "name": "my launcher",
  "type": "python",
  "request": "launch",
  "program": "/home/user/.local/bin/myProg",
  "args": [
      "install",
      "--someOption=myOption",
      "--someOtherOption=myOtherOption"
  ],
  "cwd": "${fileDirname}",
  "console": "integratedTerminal",
  "justMyCode": false
}

This config works as long as I have myInput.py (or any other file from the folder containing myInput.py) open and in focus in the editor when I click "start debugging", because I have set "cwd" to be "${fileDirname}", i.e. the folder containing the current file.

This is what I want to achieve: I want to get an error when any file other than myInput.py is in focus when I "start debugging".

These are the parts of the behavior that I want to change:

1- Currently when a file from a different folder is selected, the code runs until it reaches the part when it needs to load myInput.py, then it raises an exception and fails. What I want in this case is "not running at all", instead giving me an error, message or similar.

2- Currently when a file from the same folder as myInput.py is selected, the code runs fine. I would prefer it to "not run" and instead give me an error/message that the correct file is not selected.

Is there a way to achieve what I want (item 1 is more important)? In short, I want to have a condition based on which it is decided whether to run the program or not.

Assume that I cannot change "myProg" or how it works.

Tags:

3

Answers


  1. 0

    You can use the extension Command Variable and the command extension.commandvariable.file.fileAsKey

    If the current file does not contain /myInput.py choose a non existing command DoNotRun

    {
  "version": "2.0.0",
  "tasks": [
    {
      "name": "my launcher",
      "type": "python",
      "request": "launch",
      "program": "${input:chooseCmd}",
      "args": [
          "install",
          "--someOption=myOption",
          "--someOtherOption=myOtherOption"
      ],
      "cwd": "${fileDirname}",
      "console": "integratedTerminal",
      "justMyCode": false
    }
  ],
  "inputs": [
    {
      "id": "chooseCmd",
      "type": "command",
      "command": "extension.commandvariable.file.fileAsKey",
      "args": {
        "/myInput.py": "/home/user/.local/bin/myProg",
        "@default": "/home/user/.local/bin/DoNotRun"
      }
    }
  ]
}
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  2. 0

    If i were you i would try to specify the cwd in launch.json differently, so it looks for myInput.py file from the {workspaceFolder} instead from file dirname. Then it would look for your file from the folder you have opened in vscode not from path of the file you are currently editing.

    {
  "name": "my launcher",
  "type": "python",
  "request": "launch",
  "program": "/home/user/.local/bin/myProg",
  "args": [
      "install",
      "--someOption=myOption",
      "--someOtherOption=myOtherOption"
  ],
  "cwd": "${workspaceFolder}/path to input.py/myInput.py",
  "console": "integratedTerminal",
  "justMyCode": false
}
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  3. 0

    This sounds like an XY problem. Just set cwd in the launch config to ${workspaceFolder}/<path to directory containing myInput.py>. Even the fact that it’s "by design" that your program needs to be run from that directory might be an XY problem. But there’s not enough info to tell.

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