javascript - 无法从 promise.all() 获得响应

Question

我发现使用promise.all时无法得到响应。

下面的orders应该是一个数组

Promise.all(promises).then(function (orders) {
  console.log(orders); // nothing logged
});

完整代码

let promises = markets.map(async (mkt) => {
  return new Promise(async function (resolve, reject) {
    return functionA(paramA)
      .then(resA => {
        return res.functionB()
          .then((resB) => {
            if (resB.length > 0) {
              return functionC()
                .then(resC => {
                  console.log(resC); // successfully logged
                  return resC;
                });
            }
          });
      })
      .catch(err => console.log(err));
  });
});
Promise.all(promises).then(function (orders) {
  console.log(orders); // nothing logged
});

我该如何解决？

更新 1
我根据评论更新代码。
orders现在正在返回

(54) [undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, Array(1), undefined, undefined, undefined, undefined, undefined, undefined, undefined, Array(1), undefined, Array(1), undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined]

如何使 promise 返回没有undefined ？

const promises = markets.map(async mkt => {
  const resA = await functionA(paramA);
  const resB = await res.functionB();
  if (resB.length > 0) {
    const resC = await functionC()
    console.log(resC); // successfully logged
    return resC;
  }
});
Promise.all(promises).then(orders => {
  console.log(orders); // nothing logged
}, err => {
  console.log(err);
});

Answer 1

• Promise 构造函数忽略从其回调返回的任何值
• 你没有在任何地方打电话resolve
• 当您调用已经为您提供 Promise 的 function 时，您不需要构建 Promise

const promises = markets.map((mkt) => {
    return functionA(paramA)
        .then(resA => res.functionB())
        .then((resB) => {
            if (resB.length > 0) {
                return functionC()
                    .then(resC => {
                        console.log(resC); // successfully logged
                        return resC;
                    });
            }
        })
        .catch(handleError);
});

更好的方法是利用await使代码平坦且易于阅读。

const promises = markets.map(async (mkt) => {
    try {
        const resA = await functionA(paramA);
        const resB = await res.functionB();

        if (resB.length > 0) {
            const resC = await functionC()
            return resC;
        }
    } catch (e) {
        // handle errors
    }
});

您还可以考虑在Promise.all中处理错误，而不是在映射器中。

如果您只想包含调用functionC的项目，请在之后过滤掉空值。

Promise.all(promises).then(function (orders) {
  const filtered = orders.filter(Boolean);
  console.log(filtered);
});

Answer 2

好吧，您永远不会resolve()或reject()您正在创建的new Promise 。 避免使用Promise构造函数反模式，并且永远不要将async function作为执行者传递给new Promise ！

通过随后使用async / await而不是.then()可以大大简化此代码：

const promises = markets.map(async mkt => {
  const resA = await functionA(paramA);
  const resB = await res.functionB();
  if (resB.length > 0) {
    const resC = await functionC()
    console.log(resC); // successfully logged
    return resC;
  }
});
Promise.all(promises).then(orders => {
  console.log(orders); // nothing logged
}, err => {
  console.log(err);
});