如何让observable并行运行，并在解析时返回每个promise？

import { forkJoin, of } from 'rxjs';

// 创建三个模拟的可观察对象
const observable1 = of('Observable 1').pipe(delay(2000));
const observable2 = of('Observable 2').pipe(delay(1000));
const observable3 = of('Observable 3').pipe(delay(1500));

// 使用forkJoin操作符并行运行可观察对象
forkJoin([observable1, observable2, observable3]).subscribe(
  ([result1, result2, result3]) => {
    console.log(result1); // 输出：Observable 1
    console.log(result2); // 输出：Observable 2
    console.log(result3); // 输出：Observable 3
  },
  (error) => {
    console.error(error);
  }
);

import { forkJoin, from } from 'rxjs';

// 创建三个模拟的promise
const promise1 = new Promise((resolve) => setTimeout(() => resolve('Promise 1'), 2000));
const promise2 = new Promise((resolve) => setTimeout(() => resolve('Promise 2'), 1000));
const promise3 = new Promise((resolve) => setTimeout(() => resolve('Promise 3'), 1500));

// 使用from操作符将promise转换为可观察对象
const observable1 = from(promise1);
const observable2 = from(promise2);
const observable3 = from(promise3);

// 使用forkJoin操作符并行运行可观察对象
forkJoin([observable1, observable2, observable3]).subscribe(
  ([result1, result2, result3]) => {
    console.log(result1); // 输出：Promise 1
    console.log(result2); // 输出：Promise 2
    console.log(result3); // 输出：Promise 3
  },
  (error) => {
    console.error(error);
  }
);