javascript - Promise.all 不检查/存储所有 promise

Question

我的 promise 有问题。所有的 promise 都没有得到解决...

简短示例:

const express = require('express');
const app = express();
const router = express.Router();
const request = require('request');

router.get('/', function(req, res)
{
    var promiseArray = [];

    function promiseResolve (data) {
        return new Promise (function (resolve) {
            resolve(data);
        });
    }

    var data1 = 1;

    promiseArray.push(promiseResolve(data1));

    var data2 = 2;

    promiseArray.push(promiseResolve(data2));

    var data3 = 3;

    promiseArray.push(promiseResolve(data3));

    console.log(promiseArray);

    **// all is fine**

    **// but when we add promise in some asynchronous operation like... request for example:**

    var test = request('https://www.google.com', function (err, res, content) {
        if (res.statusCode == 200)
        {
            promiseArray.push(promiseResolve(content));
            console.log(content);
        }
    });

    Promise.all(promiseArray).then(function(data) {
        console.log(data);
    });

    **// it does not store promise in a array... I thought that at first promise.All always 'check' if all promises are stored, then execute those by resolve, reject...**

  res.render('index');
});
module.exports = router;

注释在代码中，但长话短说 - Promise.All 不会检查/存储所有 Promise。

更多关于SO“算法”的文本...以及更多...以及更多...以及更多...

Answer 1

在 Promise.all(promiseArray) 运行时，promiseArray 仅由前三个 Promise 组成 - 第四个来自 request直到请求完成后才会添加到 promiseArray 中，此时 Promise.all 已经解析。您必须为请求显式构造一个Promise，以便promiseArray在之前拥有所有必需的Promises >Promise.all 被调用:

promiseArray.push(new Promise((resolve, reject) => {
  request('https://www.google.com', function (err, res, content) {
    if (res.statusCode == 200) resolve(content);
    else reject('bad statusCode');
  });
}));