javascript - 無法從 promise.all() 獲得響應

Question

我發現使用promise.all時無法得到響應。

下面的orders應該是一個數組

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

完整代碼

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
});

我該如何解決？

更新 1
我根據評論更新代碼。
orders現在正在返回

(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 返回沒有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 構造函數忽略從其回調返回的任何值
• 你沒有在任何地方打電話resolve
• 當您調用已經為您提供 Promise 的 function 時，您不需要構建 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使代碼平坦且易於閱讀。

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
    }
});

您還可以考慮在Promise.all中處理錯誤，而不是在映射器中。

如果您只想包含調用functionC的項目，請在之后過濾掉空值。

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

Answer 2

好吧，您永遠不會resolve()或reject()您正在創建的new Promise 。 避免使用Promise構造函數反模式，並且永遠不要將async function作為執行者傳遞給new Promise ！

通過隨后使用async / await而不是.then()可以大大簡化此代碼：

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);
});