Link two positions of an object into a single position

user_16317127 0
Viewed 68

I receive in this way the object

const tests = [
        {id: 0, vencimiento: "231216000000"},
        {id: 0, monto: "99000"},
        {id: 1, vencimiento: "230117000000"},
        {id: 1, monto: "198000"},
        {id: 2, vencimiento: "230217000000"},
        {id: 2, monto: "297000"},
    ];

And I want to combine the positions with the same id

const resultado= [
     {id: 0, vencimiento: "231216000000", monto: "99000"},
     {id: 1, vencimiento: "230117000000", monto: "198000"},                             
     {id: 2, vencimiento: "230217000000", monto: "297000"},
];
javascript
3 Answers

You can use Array.prototype.reduce() and return a plain object,
then use Object.values() to get the result you wanted:

const tests = [
    {id: 0, vencimiento: "231216000000"},
    {id: 0, monto: "99000"},
    {id: 1, vencimiento: "230117000000"},
    {id: 1, monto: "198000"},
    {id: 2, vencimiento: "230217000000"},
    {id: 2, monto: "297000"},
];

const res = Object.values(tests.reduce((acc, curr) => {
  if(!acc[curr.id]) {
    acc[curr.id] = {...curr}
  } else {
    acc[curr.id] = {
      ...acc[curr.id],
      ...curr
    }
  }

  return acc;
}, {}));

console.log(res);

Array.prototype.reduce()
Object.values()

user_35616070

Try this:

const tests = [
  { id: 0, vencimiento: "231216000000" },
  { id: 0, monto: "99000" },
  { id: 1, vencimiento: "230117000000" },
  { id: 1, monto: "198000" },
  { id: 2, vencimiento: "230217000000" },
  { id: 2, monto: "297000" },
];

const newTests = []
for (let index = 0; index < tests.length; index += 2) {
  const arr = tests[index]
  const { monto } = tests[index + 1]
  newTests.push({ ...arr, monto })
}

console.log(newTests)
/* OUTPUT
[
  { id: 0, vencimiento: '231216000000', monto: '99000' },
  { id: 1, vencimiento: '230117000000', monto: '198000' },
  { id: 2, vencimiento: '230217000000', monto: '297000' }
]
*/

How it works

You can deconstruct objects by using the { } when declaring the new variable. You can reconstruct it again using the spread operator ... with arr to combine it with monto.

user_162139620

In cases where number of elements is changing reduce function is great choice

tests.reduce((p,n) => {
    if(p.map(t => t.id).includes(n.id)) {
       let i = p.findIndex(t => t.id === n.id);
       p[i] = {...p[i],...n}
       return p;
    } else {
       return [...p,n]
    }
}, [])

Result:

0: {id: 0, vencimiento: "231216000000", monto: "99000"}
1: {id: 1, vencimiento: "230117000000", monto: "198000"}
2: {id: 2, vencimiento: "230217000000", monto: "297000"}
user_63980440
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