javascript - Nodejs后台运行函数

Question

我有 Node 程序，我需要在程序开头运行两个函数稍后访问函数结果，目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 因为我稍后需要在项目中获取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 Node js 中做到这一点，如果我运行 promise.all 它会等到 getService 和 getProcess 然后去休息程序。

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，您将两个 promise 的返回值总共返回 4 秒。

当这个进程在后台运行时，您可以执行代码，当 promise 解决时会有回调，您的延迟函数将在这个阶段被调用。

检查下面的简单示例；

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();