在Windows下的VSCode中调试Python时，会忽略`.venv`

问题描述

我在Windows中有一个Python项目，该项目通过console_script提供了命令行应用程序。 Python软件包使用<project-root>/.venv中的诗歌virtualenv并以可编辑模式安装到venv中。

现在，我想在集成的VSCode终端中调试命令行应用程序。为了做到这一点，我创建了一个launch.json配置，如下所示：

{
    // 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: Device Agent (WLAN)","type": "python","request": "launch","program": "mvp-end-device-agent","args": ["<","-r","D:\\MassHunter\\Data\\demo_0000.d"],"console": "integratedTerminal"
        }
    ]
}

如果我启动调试器，则会打开一个新的终端，并且出现以下错误：

Windows PowerShell
copyright (C) Microsoft Corporation. All rights reserved.


PS D:\mvp-end-device-agent>  & 'd:\mvp-end-device-agent\.venv\Scripts\python.exe' 'c:\Users\Plasmion\.vscode\extensions\ms-python.python-2020.9.114305\pythonFiles\lib\python\debugpy\launcher' '58875' '--' 'mvp-end-device-agent' '<' '-r' 'D:\MassHunter\Data\demo_0000.d'
Traceback (most recent call last):
  File "d:\python\python386\lib\runpy.py",line 194,in _run_module_as_main
    return _run_code(code,main_globals,None,File "d:\python\python386\lib\runpy.py",line 87,in _run_code
    exec(code,run_globals)
  File "c:\Users\Plasmion\.vscode\extensions\ms-python.python-2020.9.114305\pythonFiles\lib\python\debugpy\__main__.py",line 45,in <module>
    cli.main()
  File "c:\Users\Plasmion\.vscode\extensions\ms-python.python-2020.9.114305\pythonFiles\lib\python\debugpy/..\debugpy\server\cli.py",line 430,in main
    run()
  File "c:\Users\Plasmion\.vscode\extensions\ms-python.python-2020.9.114305\pythonFiles\lib\python\debugpy/..\debugpy\server\cli.py",line 267,in run_file
    runpy.run_path(options.target,run_name=compat.force_str("__main__"))
  File "d:\python\python386\lib\runpy.py",line 264,in run_path
    code,fname = _get_code_from_file(run_name,path_name)
  File "d:\python\python386\lib\runpy.py",line 234,in _get_code_from_file
    with io.open_code(decoded_path) as f:
FileNotFoundError: [Errno 2] No such file or directory: 'D:\\mvp-end-device-agent\\mvp-end-device-agent'
PS D:\mvp-end-device-agent> & d:/mvp-end-device-agent/.venv/Scripts/Activate.ps1
(.venv) PS D:\mvp-end-device-agent>

似乎在执行命令行应用程序之前未激活venv，但是之后。根据{{​​3}}和VSCode docs - Where the extension looks for environments，诗歌venvs几乎不受ATM支持。

我已经尝试通过添加来指定microsoft/vscode-python/issues/8372

{
    "python.pythonPath": "${workspaceFolder}/.venv/Scripts/python.exe"
}

无法成功访问项目本地.vscode/settings.json文件。

我该如何解决或解决此问题？

解决方法

这不是错误。默认的终端前缀应为PS \the current working directory\，因此，当我们开始调试时，它将打开一个显示默认前缀的新终端实例。但是，当我们查看其执行脚本时：

第一个是.venv\python.exe，我们将其选择为解释器，调试完成后，环境被激活。然后再次选择调试，此终端将保持.venv处于激活状态。

,

事实证明，该问题与venv管理无关，而与必须如何处理console_script s有关。调试：https://stackoverflow.com/a/64558717/5308983