如何在VSCode Python扩展名中指定启动文件

结果发现这是 program 属性

程序-启动调试器时要运行的可执行文件或文件 https://code.visualstudio.com/docs/editor/debugging#_launchjson-attributes

{
    "name": "python launch foo",
    "type": "python",
    "request": "launch",
    "program": "${workspaceFolder}/d1/d2/foo.py",
    "cwd": "${workspaceFolder}",
    "args": [ "${env:USERNAME}", "--optionX", "x1000" ]
    "console": "integratedTerminal"
}

以前我只是用默认值复制启动配置 file 变量,该变量始终只运行工作区中打开的任何活动文件。

{
    "name": "Python: Current File",
    "type": "python",
    "request": "launch",
    "program": "${file}",
    "cwd": "${workspaceFolder}",
}