JavaScript:如何做出永不解决或拒绝的 promise

Question

我正在使用:

 await Promise.race([promise1, promise2]);

逻辑是如果 promise1 没有在 5 秒内解决/拒绝，那么 promise2 可能能够解决。所以在延迟 promise2 尝试做它的事情之后，如果它失败了，我希望 promise2 返回一个永远不会解决的 Promise ，所以它的全部直到等待 promise1。

我试过了

async function promise2(timeout=5000) {
    await new Promise(resolve => setTimeout(resolve, timeout));
    if (didStuffAndOK())  {
        return "OK"
    }
    return new Promise( () => {} )
}

return new Promise( () => {} ) 似乎被解释为 promise 被拒绝而不是从未解决。

如何做出空洞的 promise (不是在现实生活中，而是在 JavaScript 中)？

Answer 1

直接回答你的问题 new Promise( () => {} ) 永远不会解决。

这是教授:

new Promise(()=>{})
   .then(()=> console.log('promise resolved'))
   .catch(()=>console.log('promise rejected'));
console.log('FOO, so that you can see that the code was executed');

但我想你有一个不同的问题。

The logic is if promise1 has not resolved/rejected within 5s, then promise2 
might be able to resolve. So after a delay promise2 tries to do its thing, if it 
fails I wish to have promise2 return a Promise that never resolves so that its 
all up to waiting for promise1.

为此，您可以在 promise2 中解析 promise1，因为 promise 是可链接的。

(async ()=>{

  const promise1 = new Promise(()=>{}) // simulating here a very long fetch;
  
  const promise2 = new Promise((res)=>{
    
    setTimeout(()=>{
        try {
          console.log('one second passed and the fetch is still running');
          throw Error() // simulating a promise2 does it thing and fails
        } catch (e){
          res(promise1); // Chain promise1 back
        }
     }, 1000);
   });  
   await Promise.race([promise1, promise2]);
   console.log('This won\'t be printed because the previous line awaits for promise1');
})()