VS Code "python.envFile" does not seem to work with "python.defaultInterpreterPath"

user_8474894 0
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Despite specifying a python.envFile in workspace (.vscode/settings.json), python.defaultInterpreterPath does not seem to fetch the interpreter path via an environment variable, declared in the envFile.

  1. File: .env
# filename: .env
# set this in .vscode/settings.json:
# "python.envFile": "${workspaceFolder}/.env"
DEFAULT_INTERPRETER_PATH=path/to/python/interepreter
  1. File: .vscode/settings.json
// filename: .vscode/settings.json
{
    "python.envFile": "${workspaceFolder}/.env",
    "python.defaultInterpreterPath": "${env:DEFAULT_INTERPRETER_PATH}",
    "python.terminal.activateEnvironment": true,
    "python.terminal.activateEnvInCurrentTerminal": false,
    "jupyter.jupyterServerType": "local",
}

If I hard code the python.defaultInterpreterPath, it works, and auto activates the interpreter, when I open a new terminal window. But it does not activate the the interpreter from the variable (in .env file).

References

Issue opened on GitHub with VS Code

visual-studio-code environment-variables python
2 Answers

i was facing issue of module not found in vscode then i wrote the same code in pycharm which was working there properly..so u can try pycharm

user_170713990

You can use variable substitution in settings files, currently variables in environment files are not recognized.

You can also use an environment variable in the path setting using the syntax ${env:VARIABLE}. For example, if you've created a variable named PYTHON_INSTALL_LOC with a path to an interpreter, you can then use the following setting value:

"python.defaultInterpreterPath": "${env:PYTHON_INSTALL_LOC}",

Note: Variable substitution is only supported in VS Code settings files, it will not work in .env environment files.

user_191339200
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