Retry fetch request doesn't return anything

Question

This is the current code and what I've come up with:

function getToken() {
    return new Promise(async (resolve, reject) => {
        try {
            let res = await fetch(url);

            if (res.status === 418) {
                setTimeout(getToken, 1000);
            } else {
                let token = await res.text();
                console.log("1", token);
                resolve(token);
            }
        } catch(e) {
            reject(e);
        }
    });
}

async function test() {
    let token = await getToken();
    console.log("2", token);
}

test();

It logs 1 <token> but it doesn't log the other part like its supposed to (2 <token>). Is there something I'm missing or doing wrong?

`new Promise(async (` seems redundant? An `async` function will return a promise — evolutionxbox, Jun 11 '21 at 20:07
and using 418 seems mysterious to use in 2021. what's the usecase? — Bhojendra Rauniyar, Jun 11 '21 at 20:12

evolutionxbox · Accepted Answer · 2021-06-12T01:01:36.047

My very naive approach would be a mix of a "sleep" function and standard async/await syntax (no need to mix then into it).

This does not take into consideration a possible infinite loop if the URL consistently returns a 418 http code.

The biggest thing to note is that I am returning getToken() in the retry and also returning token in the else. If we don't do this token inside test will always be undefined.

function sleep(ms) {
    return new Promise(resolve => setTimeout(resolve, ms));
}

async function getToken() {
  try {
    let res = await fetch('https://httpstat.us/200');

    if (res.status === 418) {
      await sleep(1000);
      return getToken();
    } else {
      let token = await res.text();
      console.log("1", token);
      return token;
    }
  } catch (e) {}
};

async function test() {
  let token = await getToken();
  console.log("2", token);
}

test();

Nitsua · Answer 2 · 2021-06-11T22:05:14.327

0

function sleep(ms) {
    return new Promise(resolve => setTimeout(resolve, ms));
}

function get(url, retryStatusCode) {
    return new Promise(async (resolve, reject) => {
        let found = false;
        while (found !== true) {
            try {
                await sleep(1000);
                await fetch(url).then(res => { 
                    let text = res.text(); 
                    let status = res.status;
                    if (status !== retryStatusCode) {
                        found = true;
                        resolve(text);
                    }
                });
            } catch (error) {
                reject(error);
            }
        }
    });
};

Then await get(url, 404).

edited Jun 11 '21 at 22:05

answered Jun 11 '21 at 21:50

Nitsua

235
2
8

Why put an async function inside a `then`? – evolutionxbox Jun 11 '21 at 21:55
@evolutionxbox My bad, fixed in next edit – Nitsua Jun 11 '21 at 22:02

score -1 · Answer 3 · answered Jun 11 '21 at 20:29

-1

Use then

await fetch(url).then(res=>console.log(res).catch(err=>console.log(err);

I'm just logging the res here, you can do whatever you want with it

answered Jun 11 '21 at 20:29

Mohak Gupta

163
1
9

1

I think you're missing a `)` after your first console.log, and another after the second (also... given you're awaiting fetch()... there's not much reason to use .then/.catch) – Kevin B Jun 11 '21 at 20:45
Yeah,, I saw my mistaked – Mohak Gupta Jun 12 '21 at 09:41

Retry fetch request doesn't return anything

3 Answers3