在 foreach 循环中进行 api 调用并将结果推送到另一个数组中

Question

I am trying to make an api call in a for each loop. When I make an api call, the function returns the Promise. I want use that return response and check if the promise is resolved (then do something) and if rejected (then do something). But, everytime in the foreach loop, first the api makes both the calls and then goes to the then() part. What I want is in foreach loop, make an api call -> check if resolved or rejected -> do something -> make an api call -> check if resolved or rejected -> do something and so on.

//cannot change the below implementation. I need to use this already created function

    const callApiFunction = options => {
      const callback = (resolve, reject) => {
      something.call(options, resolve, reject)
    };
      return new Promise(callback);         
     }

//First attempt

const a =[];
const b =[];
list.forEach(listelement => {
    callApiFunction(options)
     .then(() => {
         a.push(listelement);
       })
     .catch(() => {
        b.push(listelement);
     });
});

//Second attempt

    const a =[];
    const b =[];
   list.forEach(listelement => {
    let promise = new Promise((resolve, reject) => {
      callApiFunction(options);
      resolve(listelement);
      reject(listelement);
    });
    promise.then((listelement) => {
      a.push(listelement);
    })
    .catch((listelement) => {
      b.push(listelement);
    });
  });

What happens in my second code block is, callApiFunction is first executed twice if list.length = 2 and then, then is executed twice. I want the callApiFunction is executed and list[0] -> then or catch executes accordingly callApiFunction is executed and list[1] -> then or catch executes accordingly

Answer 1

Does this variant with for-of loop and await work for you?

'use strict';

async function separate(list) {
  const a = [];
  const b = [];

  for (const listelement of list) {
    try {
      await callApiFunction(options);
      a.push(listelement);
    } catch (err) {
      b.push(listelement);
    }
  }

  return [a, b];
}

separate([1, 2, 3]).then(([good, bad]) => {
  console.log(`Success:\n${good.join('\n')}\n\nErrors:\n${bad.join('\n')}\n`);
});