NodeJS : While loop check the status untill sucess in fetch

Question

I have the below code which call api and get status using node-fetch . I need to use node-fetch 2.6.1 and use the code like .then-->.then cannot able to use async await .

function fetching(){
 fetch(jobsurl, requestOptions)
    .then(response => response.text())
    .then(result => {
        
        console.log("------------the job id is ------------"); 
        console.log(eval(result)[0]["id"])
        console.log("\n\n\n\n\n")
        console.log("--------------The Job Status is------------");
        console.log(eval(result)[0]["status"])
        console.log("\n\n\n\n\n")
        var job_id=eval(result)[0]["id"].toString();
        var stat=eval(result)[0]["status"];

       while(stat != "success"){
           
         fetch(jobsurl, requestOptions)
           .then(response => response.text())
           .then(result => {
               console.log("------------the job id is ------------"); 
               console.log(eval(result)[0]["id"])
               console.log("\n\n\n\n\n")
               console.log("--------------The Job Status is------------");
               console.log(eval(result)[0]["status"])
               console.log("\n\n\n\n\n")
               job_id=eval(result)[0]["id"].toString();
               stat=eval(result)[0]["status"];
            }
           // as it is success so call the main url
           fetch()....             

           }//ftehcing end

now it has stat variable inside then that give the job status success or pending if it sucess then only I can proceed. But for that I need to iterate and check when it will get success every time. so I used while loop.

But it is not working. as stat value check every time by calling fetch inside it.

Please suggest

What do you mean "cannot able to use async await"? If you can use `.then()`, you can use `async` and `await` just fine. — AKX, Apr 19 '22 at 06:55
`async` functions return promises, so if you have a non-async function you can't convert to async, you can use `.then()` there. — AKX, Apr 19 '22 at 07:10
If you're calling `fetch()` your function is already asynchronous, whether it's tagged as `async` (so you can use `await`) or not. It appears you have some basic misunderstanding of asynchronous code in nodejs. Making it `async` and using `await` will sacrifice nothing and make your function easier to program as the solution AKX shows you illustrates. There is NO reason not to do it that way. — jfriend00, Apr 19 '22 at 07:15

AKX · Accepted Answer · 2022-04-19T07:44:41.813

1

Here's an async/await reworking of your code that makes the loop pretty much trivial.

There's also

proper error handling (using response.ok)
no use of eval – response.json() should do what you want
a delay between requests (so as not to spam the server milliseconds apart)

function delay(number) {
  return new Promise(resolve => setTimeout(resolve, number));
}

async function fetching() {
  while (true) {
    const response = await fetch(jobsurl, requestOptions);
    if (!response.ok) {
      throw new Error(`HTTP error, status = ${response.status}`);
    }
    const results = await response.json();
    const {id, status} = results[0];
    console.log(`Job ${id} status: ${status}`);
    if (status === 'success') {
      break;
    }
    await delay(500);  // Wait for a bit before re-requesting
  }
  // Call main url..?
}

EDIT: If you absolutely can't use async/await, you'd need to use a recursive function, something like this, and it's pretty hard to read.

function delay(number) {
  return new Promise((resolve) => setTimeout(resolve, number));
}

function fetching() {
  function _recursive() {
    return fetch(jobsurl, requestOptions)
      .then(response => {
        if (!response.ok) {
          throw new Error(`HTTP error, status = ${response.status}`);
        }
        return response.json();
      })
      .then(results => {
        const {id, status} = results[0];
        console.log(`Job ${id} status: ${status}`);
        if (status === 'success') {
          return results;
        }
        return delay(500).then(_recursive);
      });
  }
  return _recursive();
}

function main() {
  fetching().then(results => {
    console.log(results);
    // call main URL..?
  });
}

edited Apr 19 '22 at 07:44

answered Apr 19 '22 at 07:02

AKX

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we cannot call async function fetchJobstaff it should be a sync function platform where it is executing not support this – Rajarshi Das Apr 19 '22 at 07:10
If you use `fetch()`, you're already not synchronous. – AKX Apr 19 '22 at 07:11
but my method is function fetching() not asyn function fetch..() – Rajarshi Das Apr 19 '22 at 07:12
You're using `node-fetch`, which is always asynchronous. – AKX Apr 19 '22 at 07:12
we cannot use ` const jj = await fetchJobStuff()` – Rajarshi Das Apr 19 '22 at 07:13
1

Why not? Where are you running this? You've tagged this as `node.js`, which most assuredly can do `await`. – AKX Apr 19 '22 at 07:14
Actually it is using a platform which is not supported outside declaration like above we need to define a sync method and use fetch or any async inside it and then call the method outside like `fetching()` – Rajarshi Das Apr 19 '22 at 07:19
What platform is it, _please_? In any case, `async function ...` is just syntactic sugar for making a function return a promise. If this platform doesn't support the `async function` syntax at all, you could transpile your code to plain `function`s first. – AKX Apr 19 '22 at 07:22
In any case, I added a non-async recursive reformulation. – AKX Apr 19 '22 at 07:45

NodeJS : While loop check the status untill sucess in fetch

1 Answers1