The step-by-step guide to implementing async/await specifically for a 'for loop'

Question

The step-by-step guide to implementing async/await specifically for a 'for loop'

Is there a way to make 'submitToTheOthers' function run after 'let items = []' has completed, without needing an await within 'submitToTheOthers'? I am considering using await within the for loop in 'submitToTheOthers', but I'm unsure how to do so.

If I were to use await within 'updateGeneralInfoToOther', it would mean that each server would not be updated simultaneously, and if any server encounters an error, the process would halt. Is there a better approach involving promises or another method to handle this situation?

Any suggestions on how to ensure that 'submitToTheOthers' waits before executing further actions?

PS) I am particularly interested in older syntaxes as I may need to support Internet Explorer XD XD TT

Settings.vue

<template>
  <div>
    <button class="point" @click="submitTotalServer()">submitTotalServer</button>
  </div>
</template>

... (omitted for brevity)

api/settings.js

// ...
export function updateGeneralInfoToOther(id, data, serverAddress) {
  axios.defaults.timeout = 5000;
  data.serverAddress = serverAddress
  
  return axios.post(`/api/settings/generalToOther/${id}`, data)
}
// ...

javascript vue.js

Answer 1

Answer №1

The main issue at hand is that the `for` loop in your code does not wait for the promise returned by `updateGeneralInfoToOther()` to complete. To fix this, you simply need to make sure the loop `awaits` the promise, like so:

for (let i=0; i<this.idxs.length ;i++) {
    await updateGeneralInfoToOther(1, data, this.serverFullAddress[i]).then((res) => {
        if (res.status == 200) {
            this.serverResCheck[i]= true
        } else {
            this.serverResCheck[i]= false
        }
    })
}

However, this approach has its drawbacks. You are awaiting each request separately, which can be improved. A better solution would be to execute all async operations simultaneously and then wait for them to finish together. The `submitToOthers()` function could return an array of promises instead:

return this.idx.map((_entry, i) => {
    return updateGeneralInfoToOther(1, data, this.serverFullAddress[i]).then((res) => {
        if (res.status == 200) {
            this.serverResCheck[i] = true
        } else {
            this.serverResCheck[i] = false
        }
    })
});

Then, you can utilize Promise.all() to await the resolution of this promises array:

async submitTotalServer() {
    await Promise.all(this.submitToTheOthers());

    // Additional logic goes here
    // ...
}

Answer 2

The main issue at hand is that the `for` loop in your code does not wait for the promise returned by `updateGeneralInfoToOther()` to complete. To fix this, you simply need to make sure the loop `awaits` the promise, like so:

for (let i=0; i<this.idxs.length ;i++) {
    await updateGeneralInfoToOther(1, data, this.serverFullAddress[i]).then((res) => {
        if (res.status == 200) {
            this.serverResCheck[i]= true
        } else {
            this.serverResCheck[i]= false
        }
    })
}

However, this approach has its drawbacks. You are awaiting each request separately, which can be improved. A better solution would be to execute all async operations simultaneously and then wait for them to finish together. The `submitToOthers()` function could return an array of promises instead:

return this.idx.map((_entry, i) => {
    return updateGeneralInfoToOther(1, data, this.serverFullAddress[i]).then((res) => {
        if (res.status == 200) {
            this.serverResCheck[i] = true
        } else {
            this.serverResCheck[i] = false
        }
    })
});

Then, you can utilize Promise.all() to await the resolution of this promises array:

async submitTotalServer() {
    await Promise.all(this.submitToTheOthers());

    // Additional logic goes here
    // ...
}

Answer 3

Answer №2

perhaps this could be what you're looking for for await

 async function test() {
let numbers=[1,2,3,4,5]
          for await (const element of numbers) {
           
             console.log(element);
          }
       }

Answer 4

perhaps this could be what you're looking for for await

 async function test() {
let numbers=[1,2,3,4,5]
          for await (const element of numbers) {
           
             console.log(element);
          }
       }

Answer 5

Answer №3

When you declare a function as async, you are instructing it to pause execution at any await keyword until the awaited promise is either resolved or rejected. In your submitToTheOthers function, however, there are no await calls, causing all requests to be sent simultaneously and the function to return without waiting for them to complete.

While adding an await before each call would address this issue, it would also result in sequential processing where each iteration waits for the previous one to finish. This approach might be slower compared to sending all requests concurrently, which appears feasible based on the provided code snippet.

To execute these requests in parallel, you can utilize Promise.all(), a method designed for handling multiple promises concurrently. The revised implementation below should achieve the desired outcome without requiring extensive modifications:

submitToTheOthers() {
  return Promise.all(
    this.idxs.map((_, i) => updateGeneralInfoToOther(
      1,
      { data: 'test' },
      this.serverFullAddress[i]
    ) 
      .then(r => { 
         this.serverResCheck[i] = r.status === 200;
         return r;
      })
      .catch(e => {
         this.serverResCheck[i] = false;
         return e;
      })
  ));
}

Answer 6

When you declare a function as async, you are instructing it to pause execution at any await keyword until the awaited promise is either resolved or rejected. In your submitToTheOthers function, however, there are no await calls, causing all requests to be sent simultaneously and the function to return without waiting for them to complete.

While adding an await before each call would address this issue, it would also result in sequential processing where each iteration waits for the previous one to finish. This approach might be slower compared to sending all requests concurrently, which appears feasible based on the provided code snippet.

To execute these requests in parallel, you can utilize Promise.all(), a method designed for handling multiple promises concurrently. The revised implementation below should achieve the desired outcome without requiring extensive modifications:

submitToTheOthers() {
  return Promise.all(
    this.idxs.map((_, i) => updateGeneralInfoToOther(
      1,
      { data: 'test' },
      this.serverFullAddress[i]
    ) 
      .then(r => { 
         this.serverResCheck[i] = r.status === 200;
         return r;
      })
      .catch(e => {
         this.serverResCheck[i] = false;
         return e;
      })
  ));
}

Answer 7

Answer №4

Despite labeling async submitToTheOthers, it does not actually function as an asynchronous function.

Within the function, the async line appears to be:

updateGeneralInfoToOther(1, data, this.serverFullAddress[i]')

. This function is called in a loop, but the issue arises from not using await or returning the promise.

To address this, you must include await before it, which will cause each iteration to wait for the previous one. Alternatively, you can store each promise in an array and use await Promise.all(promises) at the end of the function.

Answer 8

Despite labeling async submitToTheOthers, it does not actually function as an asynchronous function.

Within the function, the async line appears to be:

updateGeneralInfoToOther(1, data, this.serverFullAddress[i]')

. This function is called in a loop, but the issue arises from not using await or returning the promise.

To address this, you must include await before it, which will cause each iteration to wait for the previous one. Alternatively, you can store each promise in an array and use await Promise.all(promises) at the end of the function.

The step-by-step guide to implementing async/await specifically for a 'for loop'

Answer №1

Answer №2

Answer №3

Answer №4

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