JS - Do sort in an array of strings as dates

Question

I have the following array:

let datas = ["nov/2018", "set/2018", "jan/2019", "dez/2018", "out/2018"]

when I execute

datas.sort()

he orders by alphabetical order, however, I must order first by the year and then by alphabetical order.

["dez/2018", "jan/2019", "nov/2018", "out/2018", "set/2018"]

Testing today I arrived at the following line of code:

var datas = ["mar", "abr", "jan", "dez", "set", "mai", "jun", "out", "jul"];
var datas_corretas = ["jan", "fev", "mar", "abr", "mai", "jun", "jul", "ago", "set", "out", "nov", "dez"];
var result = [];

datas_corretas.forEach(function ordenar(element, index){
  var mes = datas.filter(function(valor){
  return valor == datas_corretas[index];
});
  result.push(mes[0]);
  for(element of result){
    if (element === undefined || element === null){
      result.pop(element);
    }
  }
});
console.log(result);

That way I can sort the data, but the problem is when I use it with the year

var datas = ["mar/2018", "abr/2018", "jan/2019", "dez/2018", "set/2018", "mai/2018", "jun/2018", "out/2018", "jul/2018"];

Does anyone have any idea how I can resolve this?

So for a given input, what exactly do you want the output to be? Are you actually trying to put these in date order? If so, then you'll need to sort by first comparing the year part, and then the month part. — Matt Burland
– Matt Burland, Commented Feb 15, 2019 at 19:18

Nina Scholz · Accepted Answer · 2019-02-15 19:35:24Z

1

You could split the dates and sort by year first, and then by the index of the month.

var months = ["jan", "fev", "mar", "abr", "mai", "jun", "jul", "ago", "set", "out", "nov", "dez"],
    datas = ["nov/2018", "set/2018", "jan/2019", "dez/2018", "out/2018"];

datas.sort((a, b) => {
    var aa = a.split('/'),
        bb = b.split('/');
    return aa[1] - bb[1] || months.indexOf(aa[0]) - months.indexOf(bb[0]);
});

console.log(datas);

For a faster access, you could store the months as object.

var months = { jan: 1, fev: 2, mar: 3, abr: 4, mai: 5, jun: 6, jul:7, ago: 8, set: 9, out: 10, nov: 11, dez: 12 },
    datas = ["nov/2018", "set/2018", "jan/2019", "dez/2018", "out/2018"];

datas.sort((a, b) => {
    var aa = a.split('/'),
        bb = b.split('/');
    return aa[1] - bb[1] || months[aa[0]] - months[bb[0]];
});

console.log(datas);

edited Feb 15, 2019 at 19:35

answered Feb 15, 2019 at 19:20

Nina Scholz

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Comments

guest271314 · Accepted Answer · 2019-02-15 19:18:48Z

0

however, I must order first by the year and then by alphabetical order.

["dez/2018", "jan/2019", "nov/2018", "out/2018", "set/2018"]

To achieve the expected result you can use .sort() twice, .split() at "/", .pop() and .shift()

datas.sort((a, b) => a.split("/").pop() < b.split("/").pop() ? -1 : 0)
.sort((a, b) => a.split("/").shift() < b.split("/").shift() ? -1 : 0);

answered Feb 15, 2019 at 19:18

guest271314

1

Comments

nvioli · Accepted Answer · 2019-02-15 19:28:25Z

0

TIL you can pass a custom comparator to javascript's sort function. We can use this to write our own sorting logic by first comparing the years, then the months:

let datas = ["nov/2018", "set/2018", "jan/2019", "dez/2018", "out/2018"]
const orderedMonths = ["jan", "fev", "mar", "abr", "mai", "jun", "jul", "ago", "set", "out", "nov", "dez"];

datas.sort((a,b) => {
  var [montha,yeara] = a.split('/');
  var [monthb,yearb] = b.split('/');
  
  if (yeara > yearb) return 1;
  if (yearb > yeara) return -1;
  if (orderedMonths.indexOf(montha) > orderedMonths.indexOf(monthb)) return 1;
  if (orderedMonths.indexOf(monthb) > orderedMonths.indexOf(montha)) return -1;
  return 0;
});

console.log(datas);

answered Feb 15, 2019 at 19:28

nvioli

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Comments

Alen.Toma · Accepted Answer · 2019-02-15 19:36:58Z

0

its very easy when converting the value to numbers and index.

Here se my example below.

var datas_corretas = ["jan", "fev", "mar", "abr", "mai", "jun", "jul", "ago", "set", "out", "nov", "dez"];

var datas = ["mar/2018", "abr/2018", "jan/2019", "dez/2018", "set/2018", "mai/2018", "jun/2018", "out/2018", "jul/2018"];

var result = datas.sort(function(a, b){
// index of the month
var aMonth = datas_corretas.indexOf(a.split("/")[0]);
// convert the year to number
var aYear = parseInt(a.split("/")[1]);

var bMonth = datas_corretas.indexOf(b.split("/")[0])
var bYear = parseInt(b.split("/")[1])

return aYear - bYear || aMonth - bMonth;
});

console.log(result)

answered Feb 15, 2019 at 19:36

Alen.Toma

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Comments

RobG · Accepted Answer · 2019-02-18 00:44:44Z

0

I must order first by the year and then by alphabetical order.

Answers seem to all sort the month chronologically, so here's how to sort them alphabetically. Split in to month and year, then firstly compare years then compare months using localeCompare.

let datas = ["nov/2018", "set/2018", "jan/2019", "dez/2018", "out/2018"]

datas.sort((a, b) => {
  let [amon, ayr] = a.split('/');
  let [bmon, byr] = b.split('/');
  return (ayr - byr) || amon.localeCompare(bmon);
});

console.log(datas);

answered Feb 18, 2019 at 0:44

RobG

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