为什么 promise 用 undefined 解决? [英] Why is promise resolving with undefined?

问题描述

var firstPromise = new Promise((resolve, reject) => {
  resolve('first promise');
});

firstPromise.then(() => {
  return new Promise((resolve, reject) => {
    resolve('second promise');
  }).then((result) => {
    console.log('hello');
  });
}).then((result) => {
  console.log(result);
});

我知道这不是编写此承诺链的最佳方式，但我想知道为什么最后一个 .then 会执行.我没有用 console.log('hello') 返回任何东西，那么第二个承诺的 .then 不会永远解决吗?

I know this is not the best way to write this promise chain, but I was wondering why the last .then executes at all. I'm not returning anything with console.log('hello'), so wouldn't the .then off of the second promise never resolve?

推荐答案

因为您已经将多个 promise 链接在一起，并且您的 .then() 处理程序之一不返回任何内容.

Because you've chained several promises together and one of your .then() handlers returns nothing.

这部分:

.then((result) => {
    console.log('hello');
    // since there is no return value here, 
    // the promise chain's resolved value becomes undefined
});

什么都不返回，本质上与return undefined相同，因此链的解析值变为undefined.

returns nothing which is essentially the same as return undefined and therefore the resolved value of the chain becomes undefined.

您可以将其更改为这样以保留已解析的值:

You can change it to this to preserve the resolved value:

.then((result) => {
    console.log('hello');
    return result;         // preserve resolved value of the promise chain
});

请记住，每个 .then() 处理程序的返回值都会成为链的解析值.没有返回值使得解析值成为undefined.

Remember that the return value of every .then() handler becomes the resolved value of the chain going forward. No return value makes the resolved value be undefined.

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