承诺值和 return 承诺 JS 的总和

Question

我有一个任务是创建一个函数 summarize1 来接收 promise 和 returns promise，它们的值之和如下所示：

const promise1 = Promise.resolve(4);
const promise2 = new Promise((resolve) => resolve(2));
summarize1(promise1, promise2).then((sum) => {console.log(sum);}); // result: 6

通过此测试实施：

describe('summarize1', () => {
    it('should return 9 for given promises', async () => {
        const promise1 = Promise.resolve(9);
        const res = await summarize1(promise1);
        assert.strictEqual(res, 9);
    });

    it('should return 6 for given promises', async () => {
        const promise1 = Promise.resolve(2);
        const promise2 = Promise.resolve(4);
        const res = await summarize1(promise1, promise2);
        assert.strictEqual(res, 6);
    });

    it('should return 11 for given promises', async () => {
        const promise1 = Promise.resolve(3);
        const promise2 = new Promise(resolve => resolve(9));
        const res = await summarize1(promise1, promise2);
        assert.strictEqual(res, 12);
    });

    it('should return 16 for given promises', async () => {
        const promise1 = Promise.resolve(10);
        const promise2 = new Promise(resolve => resolve(5));
        const promise3 = new Promise(resolve => {setTimeout(() => {resolve(1)}, 300)});
        const res = await summarize1(promise1, promise2, promise3);
        assert.strictEqual(res, 16);
    });
});

还有一个名为 summarize2 的 async 函数，它接收 promise 和 returns promise，它们的值之和如上图所示总结1。通过此测试：

describe('summarize2', () => {
    it('should return 9 for given promises', async () => {
        const promise1 = Promise.resolve(9);
        const res = await summarize2(promise1);
        assert.strictEqual(res, 9);
    });

    it('should return 6 for given promises', async () => {
        const promise1 = Promise.resolve(2);
        const promise2 = Promise.resolve(4);
        const res = await summarize2(promise1, promise2);
        assert.strictEqual(res, 6);
    });

    it('should return 11 for given promises', async () => {
        const promise1 = Promise.resolve(3);
        const promise2 = new Promise(resolve => resolve(9));
        const res = await summarize2(promise1, promise2);
        assert.strictEqual(res, 12);
    });

    it('should return 16 for given promises', async () => {
        const promise1 = Promise.resolve(10);
        const promise2 = new Promise(resolve => resolve(5));
        const promise3 = new Promise(resolve => {setTimeout(() => {resolve(1)}, 300)});
        const res = await summarize2(promise1, promise2, promise3);
        assert.strictEqual(res, 16);
    });

    it('should return 23 for given promises', async () => {
        const promise1 = Promise.resolve(11);
        const promise2 = new Promise(resolve => resolve(7));
        const promise3 = new Promise(resolve => {setTimeout(() => {resolve(5)}, 500)});
        const res = await summarize2(promise1, promise2, promise3);
        assert.strictEqual(res, 23);
    });

    it('should return 16 for given promises', async () => {
        const promise1 = new Promise(resolve => {setTimeout(() => {resolve(1)}, 500)});
        const promise2 = new Promise(resolve => {setTimeout(() => {resolve(3)}, 500)});
        const promise3 = new Promise(resolve => {setTimeout(() => {resolve(5)}, 500)});
        const promise4 = new Promise(resolve => {setTimeout(() => {resolve(7)}, 500)});
        const res = await summarize2(promise1, promise2, promise3, promise4);
        assert.strictEqual(res, 16);
    });

    it('should return 42 for given promises', async () => {
        const promise1 = Promise.resolve(21);
        const promise2 = Promise.resolve(21);
        const res = await summarize2(promise1, promise2);
        assert.strictEqual(res, 42);
    });
});

问题是我不明白如何实现这个任务来满足条件，能够创建我们想要的尽可能多的承诺，然后只是从 resolve 中获得它们的值的总和。

我尝试了 4 次，但没有任何效果，我不知道该怎么做，我搜索了每份文档、每份参考文献，但没有找到我需要的东西。 PS：我是 JS 新手

这是我的第一次尝试：

    new Promise((resolve) => resolve(value));
    function sum(value){
     return new Promise((resolve)=>{
    resolve(value+value);
    });
    }

第二次尝试：

async function getFirstPromise(){
   
        return new Promise((resolve)=>resolve(value1));
    }
    async function getSecondPromise(){
    
        return new Promise((resolve)=>resolve(value2));
    }
   
    const sum = new Promise((resolve)=>{
        var promise1 = await getFirstPromise();
        var promise2 = await getSecondPromise();
        resolve(promise1+promise2);
    })
    return sum.then(function(sum){
        console.log(sum);
    });

第三次尝试：

const promise1 = new Promise(function (resolve) {
        setTimeout(() => {
            resolve(value);
        }, 1000);
    });
    const promise2 = new Promise(function (resolve) {
        setTimeout(() => {
            resolve(value);
        }, 1000);
    });
    const sum = new Promise(function (resolve) {
        setTimeout(() => {
            resolve(promise1 + promise2);
        }, 1000);
        
    });
    sum.then(function () {
        return sum;
    });

最后一个：

   let promise1;
    let promise2;
    let sum = new Promise((resolve)=>{
resolve(sum);
    });
    sum.then((value)=>{
        return value;
    })

也许在某个地方我走对了路，但由于缺乏经验，我不知道我应该做什么以及提前expect.Please help.Thanks

Answer 1

您遗漏了一些 JavaScript 可以做到的关键部分

使用rest parameters
等待多个承诺解决，然后使用Promise.all

const summarize1 = (...promises) => {
  return Promise.all(promises).then(nums => {
    let sum = 0;
    for (const num of nums) {
      sum += num;
    }
    return sum;
  })
}

const promise1 = Promise.resolve(4);
const promise2 = new Promise((resolve) => resolve(2));
summarize1(promise1, promise2).then((sum) => {
  console.log(sum);
});

既然你提到你是 JS 的新手，我没有使用任何不必要的东西来解释上面的解决方案。但是，数组上有一个实用函数，称为 reduce，可以用来代替上面的 for 循环

const summarize1 = (...promises) => {
  return Promise.all(promises).then(nums => nums.reduce((sum, ele) => sum + ele));
}

const promise1 = Promise.resolve(4);
const promise2 = new Promise((resolve) => resolve(2));
summarize1(promise1, promise2).then((sum) => {
  console.log(sum);
});

承诺值和 return 承诺 JS 的总和

Sum of promise values and return promise JS

javascript

asynchronous

task

promise