I am using:
await Promise.race([promise1, promise2]);
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.
I tried
async function promise2(timeout=5000) {
await new Promise(resolve => setTimeout(resolve, timeout));
if (didStuffAndOK()) {
return "OK"
}
return new Promise( () => {} )
}
return new Promise( () => {} ) seems to be interpreted as the promise being rejected rather than never resolved.
How does one make an empty promise (not in real life, in JavaScript)?
To answer straight your question new Promise( () => {} ) never resolves.
Here is the prof :
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');But I guess you have a diffrent question.
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.
For this you can resolve in the promise2 the promise1, because promises are chainable.
(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');
})()