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0

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subtract same variables in single object javascript

const token = [
  {"token":"d2r4Z62OTGiPyNmdHTUfny",
  "time":1652767811},
  {"token":"dnl13twkQIqifdvaxp1t6e",
  "time":1652767811},
  {"token":"eDZxQu0FSWm72D2-T1md5X",
  "time":1652767811},
  {"token":"dnl13twkQIqifdvaxp1t6e",
  "time":1652767811}]; // Try edit me
  const arr = [];
  for (var i=0; i<=token.length-1; i++)
  {
    const millis = Date.now();
    const time  = Math.floor(millis / 1000);
    if (token[i].time > time) {
      arr.push(token[i].token)
    }
   }
   console.log(arr);

between these two token variables are the same, I want to distinguish them from each other, how can I do that? Example Output:

const example = [
  "d2r4Z62OTGiPyNmdHTUfny",
  "eDZxQu0FSWm72D2-T1md5X",
  "dnl13twkQIqifdvaxp1t6e"
];
console.log(example);

about 4 years ago · Juan Pablo Isaza
2 answers
Answer question

0

The easiest way with your current logic is that you can use includes to check token with the arr results. If it's in arr, we don't need to push it to arr.

const token = [
  {"token":"d2r4Z62OTGiPyNmdHTUfny",
  "time":1652767811},
  {"token":"dnl13twkQIqifdvaxp1t6e",
  "time":1652767811},
  {"token":"eDZxQu0FSWm72D2-T1md5X",
  "time":1652767811},
  {"token":"dnl13twkQIqifdvaxp1t6e",
  "time":1652767811}];
  const arr = [];
  for (var i=0; i<=token.length-1; i++)
  {
    const millis = Date.now();
    const time  = Math.floor(millis / 1000);
    if (!arr.includes(token[i].token) && token[i].time > time) {
      arr.push(token[i].token)
    }
   }
   console.log(arr);

about 4 years ago · Juan Pablo Isaza Report

0

Just use a Set as it doesn't allow for duplicates:

const tokens = [
  {"token":"d2r4Z62OTGiPyNmdHTUfny", "time":1652767811},
  {"token":"dnl13twkQIqifdvaxp1t6e", "time":1652767811},
  {"token":"eDZxQu0FSWm72D2-T1md5X", "time":1652767811},
  {"token":"dnl13twkQIqifdvaxp1t6e", "time":1652767811}
];

const now = Math.floor(Date.now() / 1000);
const result = new Set(
  tokens.filter(({ time }) => time > now).map(({ token }) => token)
);

console.log(...result);

about 4 years ago · Juan Pablo Isaza Report
Answer question
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