Convert the local time to another time zone with this JavaScript

Question

I need to increase the current date in Moscow time by 1 day every time the clock is 15:00 and more. I did it at local time, but I can’t do it according to Moscow time (UTC + 3)

  function date() {
  const today = new Date();
  const t = today.getHours();
  const dtfRU = new Intl.DateTimeFormat('ru', {
    month: 'long', day: '2-digit',
  });

  if (t >= 15) {
    today.setDate(today.getDate() + 1);
    document.querySelector('.date').innerHTML = dtfRU.format(today);
  } else document.querySelector('.date').innerHTML = dtfRU.format(today);

}

document.addEventListener("DOMContentLoaded", date);

Does this answer your question? [Add days to JavaScript Date](https://stackoverflow.com/questions/563406/add-days-to-javascript-date) — Jose Manuel Gomez Alvarez, Jun 18 '20 at 18:13

Matt Johnson-Pint · Answer 1 · 2020-06-19T16:00:19.597

0

You can retrieve the local hour in Moscow as follows:

// Get a DateTimeFormat object for the hour in Moscow in 24-hour format
const dtf = Intl.DateTimeFormat('en', {
  timeZone: 'Europe/Moscow',
  hour: 'numeric',
  hour12: false
});

// The above will create a format that has only the hour, so you can just use it.
const hour = +dtf.format();

console.log("hour:", hour);

Alternatively, if you decide you need more than just the hour, use formatToParts. For example:

const dtf = Intl.DateTimeFormat('en', {
  timeZone: 'Europe/Moscow',
  hour: 'numeric',
  hour12: false,
  minute: 'numeric'
});

const parts = dtf.formatToParts();
const hour = +parts.find(x => x.type === 'hour').value;
const minute = +parts.find(x => x.type === 'minute').value;

console.log("hour:", hour);
console.log("minute:", minute);

You can then use that in the rest of your code however you wish.

edited Jun 19 '20 at 16:00

answered Jun 18 '20 at 18:12

Matt Johnson-Pint

230,703
74
448
575

1

It's simpler to use *format* rather than *formatToParts* as it returns just the hour (e.g. "01") instead of an array of objects that you then need to dig into. :-) – RobG Jun 18 '20 at 22:29
I suppose, since there's just one part. I use this often when there are multiple parts, but then I also examine the `type` of each part to pick the right one. – Matt Johnson-Pint Jun 19 '20 at 00:51
Yes, I should have said "*in this case…*". I usually reduce the result of *formatToParts* to an object of type:part, tossing out (or just ignoring) all the literals. – RobG Jun 19 '20 at 03:41

score 0 · Answer 2 · answered Jun 19 '20 at 07:35

I found a solution here: enter link description here I needed to do something like this:

function calcTime() {
  const now = new Date();
  const utc = now.getTime() + (now.getTimezoneOffset() * 60000);

  const d = new Date(utc + (3600000 * 3));
  const h = d.getHours();
  const dtfRU = new Intl.DateTimeFormat('ru', {
    month: 'long', day: '2-digit',
  });

  if (h >= 15) {
    const newd = new Date(utc + (3600000 * 3 * 9));
    document.querySelector('.date').innerHTML = dtfRU.format(newd);
  } else document.querySelector('.date').innerHTML = dtfRU.format(d);
}
document.addEventListener("DOMContentLoaded", calcTime);

Convert the local time to another time zone with this JavaScript

2 Answers2