JavaScript：如何制作永不解决或拒绝的 promise

Question

我在用：

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

逻辑是如果promise1在5s 内没有解决/拒绝，那么promise2可能能够解决。 因此，在延迟之后promise2尝试做它的事情，如果它失败了，我希望promise2返回一个永远不会解决的Promise ，以便它完全等待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。

如何制作一个空的 promise（不是在现实生活中，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中解决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'); })()