如何在VSCode中为不同Python文件映射启动配置，实现一键快捷运行？

要让VSCode根据当前打开的Python文件自动匹配启动配置，只需给现有两个配置添加条件判断规则，修改.vscode/launch.json如下：

{
    "version": "0.2.0",
    "configurations": [
        {
            "name": "Python: app.py",
            "type": "python",
            "request": "launch",
            "module": "streamlit",
            "console": "integratedTerminal",
            "justMyCode": true,
            "args": [
                "run",
                "app.py"
            ],
            "serverReadyAction": {
                "pattern": "You can now view your Streamlit app in your browser",
                "uriFormat": "http://localhost:8501",
                "action": "openExternally"
            },
            // 仅当前文件为app.py时触发该配置
            "condition": "equals('${fileBasename}', 'app.py')"
        },
        {
            "name": "Python: Current File",
            "type": "python",
            "request": "launch",
            "program": "${file}",
            "console": "integratedTerminal",
            "justMyCode": true,
            // 当前文件非app.py时触发该配置
            "condition": "notEquals('${fileBasename}', 'app.py')"
        }
    ]
}

• 打开app.py后按F5或Ctrl+F5，自动用Streamlit启动应用
• 打开其他.py文件时，快捷键会自动触发普通Python文件运行配置
• 全程无需手动选择启动配置，完全自动匹配当前文件类型

内容的提问来源于stack exchange，提问作者Mikalai Trafimovich