Question

我正在使用一些Promise（）函数来获取一些数据，但我在项目中陷入了这个问题。

example1 = () => new Promise(function(resolve, reject) {
  setTimeout(function() {
    resolve('foo1');
  }, 3000);
});

example2 = () => new Promise(function(resolve, reject) {
  setTimeout(function() {
    resolve('foo2');
  }, 3000);
});

doStuff = () =>  {
  const listExample = ['a','b','c'];
  let s = "";
  listExample.forEach((item,index) => {
    console.log(item);
    example1().then(() => {
        console.log("Fisrt");
        s = item;
    });
    example2().then(() => {
        console.log("Second");
    });
  });
  console.log("The End");
};

如果我在代码上调用doStuff函数，则结果不正确，下面显示了我期望的结果。

RESULT                EXPECTED
a                     a
b                     First
c                     Second
The End               b
Fisrt                 First
Second                Second
Fisrt                 c
Second                First
Fisrt                 Second
Second                The End

在函数结尾，无论我如何尝试，变量s都将返回为“”，我希望s为“ c”。

Answer 1

听起来好像您希望每个Promise都先等待解决，然后再初始化下一个：您可以通过await分别Promise来完成此操作放在async函数中（并且您必须使用标准的for循环与await进行异步迭代）：

const example1 = () => new Promise(function(resolve, reject) {
  setTimeout(function() {
    resolve('foo1');
  }, 500);
});

const example2 = () => new Promise(function(resolve, reject) {
  setTimeout(function() {
    resolve('foo2');
  }, 500);
});

const doStuff = async () =>  {
  const listExample = ['a','b','c'];
  for (let i = 0; i < listExample.length; i++) {
    console.log(listExample[i]);
    await example1();
    const s = listExample[i];
    console.log("Fisrt");
    await example2();
    console.log("Second");
  }
  console.log("The End");
};

doStuff();

await仅是Promise s的语法糖-它可能（一目了然很难阅读），而无需使用{{1} } / async：

await

Answer 2

如果您不想在开始下一个承诺之前不等待每个诺言完成；

在您所有的诺言都解决之后，您可以使用Promise.all（）运行某些内容；

example1 = () => new Promise(function(resolve, reject) {
  setTimeout(function() {
    resolve('foo1');
  }, 3000);
});

example2 = () => new Promise(function(resolve, reject) {
  setTimeout(function() {
    resolve('foo2');
  }, 3000);
});

doStuff = () => {
  const listExample = ['a','b','c'];
  let s = "";
  let promises = []; // hold all the promises
  listExample.forEach((item,index) => {
    s = item; //moved
    promises.push(example1() //add each promise to the array
    .then(() => {  
        console.log(item); //moved
        console.log("First");
    }));
    promises.push(example2() //add each promise to the array
    .then(() => {
        console.log("Second");
    }));
  });
  Promise.all(promises) //wait for all the promises to finish (returns a promise)
  .then(() => console.log("The End")); 
  return s;
};
doStuff();

如何在forEach循环中等待Promise

2 个答案: