Node.js：停止执行除单个功能以外的所有代码吗？

Question

Is it possible to stop execution of all Node.js code, except for a single async function?

Kind of like running process.exit() , but process.exit() allows only sync functions.

Ie a way to exit the current call stack, clear the message queue, unregister all event handlers, and then run a single async function?

Answer 1

This is an interested use case... one thing you could do is fork a detached child_process that handles the async clean up for you. This way, you can send a message to the child prompting it to perform any cleanup prior to triggering a process.exit() on the parent to stop its process. The child should exit naturally through an emptying of its own event loop, but I put another process.exit() in the cleanup code just in case:

app.js

const { fork } = require('child_process');
const subProcess = fork(`${__dirname}/cleanup.js`, {
  detached: true
});

//ensure parent can exit independent of child
subProcess.unref();

setTimeout(() => {
  //do some standard node.js, async stuff
  //all done... send clean up message and exit parent
  subProcess.send({ exit: true });
  //disconnect to sever any further communication
  subProcess.disconnect()
  process.exit(1);
}, 5000)

cleanup.js

asyncCleanUp();

async function asyncCleanUp(){
  setTimeout(() => {
    console.log('async cleanup completed... exiting.')
  }, 5000)
  return 'done'
}

console.log('cleanup script forked');
process.on('message', async (message) => {
  console.log(message)
  if(message.exit) {
    const result = await asyncCleanUp();
    console.log(result);
    process.exit(1);
  }
});