我想在nodejs创建子进程以调试python代码后立即跳入python代码
Server.js:
const spawn = require("child_process").spawn;
const pythonProcess = spawn('python',["MyPythonFile.py"]);
console.log("Finished!")
MyPythonFile.py:
print("First line I'd like to debug")
print("Second line I'd like to debug")
print("Third line I'd like to debug")
Server.js和MyPythonFile.py都位于同一目录中 有什么办法告诉Visual Studio代码将python文件附加到调试器或为python代码创建新的调试器实例?