python - 如何在 VSCode Python 扩展中指定启动文件

Question

在 Visual Studio 的 Python Tools for Visual Studio 中(url)当你有一个 Python 项目文件时，会有一个 Startup File 的概念。 .

Each Python project has one assigned start-up file, shown in boldface in Solution Explorer. The startup file is the file that's run when you start debugging (F5 or Debug > Start Debugging) or when you run your project in the Interactive window (Shift+Alt+F5 or Debug > Execute Project in Python Interactive). To change it, right-click the new file and select Set as Startup File.

VSCode 中的等价物是什么？的 Python 扩展？如何定位我目录中的特定文件以使用从我的 launch.json 中选择的当前调试设置运行文件

Answer 1

原来是 program属性。

program - executable or file to run when launching the debuggerhttps://code.visualstudio.com/docs/editor/debugging#_launchjson-attributes

举个例子...

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

以前我只是用默认的 file 复制启动配置变量，它总是只运行工作区中打开的任何事件文件。

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