javascript - Promise.all 忽略一个 Promise 并调用另一个 Promise 两次

Question

我正在尝试使用 Promise.all 运行几个网络调用，然后在它们全部完成时告诉我。在我的例子中，我有三个 promise 。这些 promise 是从不同的对象返回的，但其中两个具有相同的方法签名。第一个 promise 从未运行，但第二个 promise 被调用两次。如果我更改其中一种方法的名称，它会起作用，但参数的数量似乎并不重要。代码示例:

const test = {};

test.ObjectOne = () => {
  aMethod = (paramOne, paramTwo) => new Promise((resolve, reject) => {
    resolve("Method one");
  });
  
  bMethod = (paramOne, paramTwo) => new Promise((resolve, reject) => {
    resolve("Method one two");
  });

  return this;
};

test.ObjectTwo = () => {
  aMethod = (paramOne, paramTwo) => new Promise((resolve, reject) => {
    resolve("Method two");
  });

  return this;
};

const objectOne = test.ObjectOne();
const objectTwo = test.ObjectTwo();

Promise.all([
  objectOne.aMethod(1, 2),
  objectOne.bMethod(1, 3),
  objectTwo.aMethod(1, 2),
]).then((responses) => {
  console.log(responses);
});

我希望输出是:

["Method one", "Method one two", "Method two"]

但实际输出是:

["Method two", "Method one two", "Method two"]

我看到的问题是什么？

致以诚挚的问候

Answer 1

您没有正确定义对象。这:

test.ObjectTwo = () => {
  aMethod = (paramOne, paramTwo) => new Promise((resolve, reject) => {
    resolve("Method two");
  });

  return this;
};

创建一个全局变量aMethod(这会覆盖之前定义的aMethod)并返回全局对象(window)。

这是可以做到的:

const test = {};

test.ObjectOne = () => ({
    aMethod: (paramOne, paramTwo) => new Promise((resolve, reject) => {
        resolve("Method one");
    }),
    bMethod: (paramOne, paramTwo) => new Promise((resolve, reject) => {
        resolve("Method one two");
    })
});

test.ObjectTwo = () => ({
    aMethod: (paramOne, paramTwo) => new Promise((resolve, reject) => {
        resolve("Method two");
    })
});

const objectOne = test.ObjectOne();
const objectTwo = test.ObjectTwo();

Promise.all([
    objectOne.aMethod(1, 2),
    objectOne.bMethod(1, 3),
    objectTwo.aMethod(1, 2),
]).then((responses) => {
    console.log(responses);
});