React-native Detecting the first successful fetch - node.js

Question

in React-Native I have a bunch of IPs that I try to fetch simultaneously. The first one to answer with a specific status code is the one I am looking for. This part happens at the app startup, so it needs to be as fast as possible. Using the async library, my code is like this:
// Here an array with a bunch of IPs
// ...
async.detect(uris, function(uri, callback) {
// Fetching a specific URL associated with the IP
fetch(`http://${uri}/productionservice/DataService.svc/`)
.then((response) => {
// If the URL answers with a 401 status code I know it's the one I'm looking for
if(response.status == '401') {
callback(null, true);
// Otherwise It's not
} else {
callback(null, false)
}
})
.catch((error) => {
callback(null, false)
});
}, function(err, result) {
if(typeof(result)=='undefined') {
console.log('No result found');
}
console.log(result);
});
}
However, when one of the tests succeeds, I do get a result, but when none succeeds, the detect method hangs indefinitely, never letting me know that none of the IPs is returning the answer I expect.
My question is: How, using async.detect and RN's fetch, can I fetch multiple links, get a result if my test succeeds, or a false statement if none of them succeeds.
Thank you.

Answer 1

Using async await you can do something along these lines:
async function detect(uris) {
const promises = [];
uris.forEach((uri) => promises.push(fetch(`http://${uri}/productionservice/DataService.svc/`)));
const responses = await Promise.all(promises);
for (let i = 0; i < responses.length; i++) {
if (responses[i] && responses[i].status === '401') {
return true;
}
}
return false;
}