Nodejs在后台运行功能

Question

我有节点程序，我需要在程序开头运行两个函数 稍后访问函数结果，当前每次等待每个函数都有效， 但是，为了节省时间而不是等待GetService 和GetProcess，因为我稍后需要项目中的数据 获取此数据大约需要 4 秒，我想在后台运行它，因为我不需要立即获得结果， 我如何在节点 js 中做到这一点，如果我运行 promise.all 它会等到 getService 和 getProcess 然后去程序的其余部分。

一个例子

function main() {

//I want to run this both function in background to save time
let service = await GetServices();
this.process = await GetProcess();


…..//Here additional code is running 

//let say that after 30 second this code is called
 Let users = GetUser(service);

 Let users = GetAdress(this.process);
} 

我实际上正在运行 yeoman 生成器 https://yeoman.io/authoring/ https://yeoman.io/authoring/user-interactions.html

export default class myGenerator extends Generator {

//here I want run those function in background to save time as the prompt to the user takes some time (lets say user have many questions...) 
async initializing() {
    let service = await GetServices();
    this.process = await GetProcess();
}

async prompting() {
    const answers = await this.prompt([
      {
        type: "input",
        name: "name",
        message: "Your project name",
        default: this.appname // Default to current folder name
      },
      {
        type: "confirm",
        name: "list",
        choises: this.process //here I need to data from the function running in background
      }
    ]);

}

Answer 1

假设getServices() 可能需要 3 秒，getProcess() 可能需要 4 秒，所以如果您同时运行这两个函数，您将在总共 4 秒内返回两个 Promise 的返回值。

你可以在这个进程在后台运行的时候执行代码，当promise解决时会有一个回调，你的后期函数会在这个阶段被调用。

查看下面的简单示例；

let service;
let process;

function main() {

    // Both functions will execute in background
    Promise.all([getServices(), getProcess()]).then((val) => {

        service = val[0];
        process = val[1];

        console.log(service, process);

        // Aafter completed this code will be called
        // let users = GetUser(service);
        // let users = GetAdress(process);
        console.log('I am called after all promises completed.')
    });

    // Current example.
    // let service = await GetServices();
    // this.process = await GetProcess();

    /* Code blocks.. */
    console.log('Code will execute without delay...')
}

function getServices() {
    return new Promise((resolve, reject) => {
        setTimeout(() => {
            resolve("service is returned")
        }, 3000);
    });
}

function getProcess() {
    return new Promise((resolve, reject) => {
        setTimeout(() => {
            resolve("process is returned")
        }, 4000);
    });
}

main();

Answer 2

要做你需要的人，你必须了解the event loop。

Nodejs 被设计为在单个线程中工作，这与 Go 等语言不同，但是 nodejs 在不同线程上处理进程。所以你可以使用nextTick()向主线程添加一个新事件，它将在整个块的末尾执行。

    function main() {

    //I want to run this both function in background to save time
    let service = await GetServices();
    this.process = await GetProcess();


    …..//Here additional code is running 

    //Let say that after 30 second this code is called
     Let users = GetUser(service);

     Let users = GetAdr(this.process);
    } 


 function someFunction(){
 // do something...
  }
 main();
 process.nextTick(someFunction());// happens after all main () processes are terminated...

Answer 3

您可以启动异步操作但还没有等待它：

function suppressUnhandledRejections(p) {
  p.catch(() => {});
  return p;
}

async function main() {
  // We have to suppress unhandled rejections on these promises. If they become
  // rejected before we await them later, we'd get a warning otherwise.
  const servicePromise = suppressUnhandledRejections(GetServices());
  this.processPromise = suppressUnhandledRejections(GetProcess());

  // Do other stuff
  const service = await servicePromise;
  const process = await this.processPromise;
}

还可以考虑使用Promise.all()，它返回一个完成所有传递给它的承诺的承诺。

async function main() {
  const [ services, process, somethingElse ] = await Promise.all([
    GetServices(),
    GetProcess(),
    SomeOtherAsyncOperation(),
  ]);

  // Use the results.
}