Using JavaScript to find the weekday of the same date next year

Question

Using JavaScript to find the weekday of the same date next year

Struggling to determine the weekday of a particular date next year? Take for example Tuesday, April 19, 2016 as the given date. What we need to calculate is: TUESDAY, April 18, 2017. Consistency with weekdays is crucial in this scenario.

The challenge lies in handling leap years. Help on this matter would be greatly appreciated.

$('.changeYear').click(function() {
        var appointments = [];
        for (var i = 1; i <= 22; i++) {
            var date = $('#tr-fix-tag-' + i).val();
            if (date !== "") {
                var splittedDate = date.substr(5).split('.');
                appointments.push(new Date(splittedDate[2], splittedDate[1] - 1, splittedDate[0]));
            }
        }

        var counter = 1;
        $.each(appointments, function() {
            var field = $('#tr-fix-tag-' + counter);
            var date = $(this)[0];
            var newDate = new Date(date.setFullYear(date.getFullYear() + 1));

            if (isSchaltjahr(newDate.getFullYear()))
                field.val(formatDate(new Date(newDate.setDate(newDate.getDate() - 2))));
            else
                field.val(formatDate(new Date(newDate.setDate(newDate.getDate() - 1))));

            counter++;
        });
    });

javascript date leap-year

Answer 1

Answer №1

To keep the day the same when adding a year, you can adjust the resulting date by finding the nearest date with the same day of the week.

function sameDayNextYear(d) {
  // Duplicate the date
  var t = new Date(+d);
  // Get the current day number
  var startDayNum = d.getDay();
  // Add one year
  t.setFullYear(t.getFullYear() + 1);
  // Find the closest date with the same day number
  var diff = startDayNum - t.getDay();
  // If the difference is more than 3 days, subtract instead of add
  diff = (diff > 3) ? diff-7 : diff;
  t.setDate(t.getDate() + diff);
  return t;  
}

[new Date(2016, 1, 13), 
 new Date(2016, 1, 5),   
 new Date(2016, 1, 6),   
 new Date(2016, 1, 29),  
 new Date(2016, 2, 1)]   
 .forEach(function(d) {
  document.write(d + '<br>' + sameDayNextYear(d) + '<br><br>');
});

This method works for any day in any year, whether it's a leap year or not.

However, in some cases the resulting date may fall into the next or previous month (e.g. 2016-03-01 becomes 2017-02-28). To avoid this, you can compare the start and end months and adjust by a week to stay within the original month (for example, go back to 2017-03-07).

Answer 2

To keep the day the same when adding a year, you can adjust the resulting date by finding the nearest date with the same day of the week.

function sameDayNextYear(d) {
  // Duplicate the date
  var t = new Date(+d);
  // Get the current day number
  var startDayNum = d.getDay();
  // Add one year
  t.setFullYear(t.getFullYear() + 1);
  // Find the closest date with the same day number
  var diff = startDayNum - t.getDay();
  // If the difference is more than 3 days, subtract instead of add
  diff = (diff > 3) ? diff-7 : diff;
  t.setDate(t.getDate() + diff);
  return t;  
}

[new Date(2016, 1, 13), 
 new Date(2016, 1, 5),   
 new Date(2016, 1, 6),   
 new Date(2016, 1, 29),  
 new Date(2016, 2, 1)]   
 .forEach(function(d) {
  document.write(d + '<br>' + sameDayNextYear(d) + '<br><br>');
});

This method works for any day in any year, whether it's a leap year or not.

However, in some cases the resulting date may fall into the next or previous month (e.g. 2016-03-01 becomes 2017-02-28). To avoid this, you can compare the start and end months and adjust by a week to stay within the original month (for example, go back to 2017-03-07).

Answer 3

Answer №2

If I were to provide guidance, I would suggest following this code snippet and proceeding towards the desired date:

let DaysOfWeek=[
    "Sunday",
    "Monday",
    "Tuesday",
    "Wednesday",
    "Thursday",
    "Friday",
    "Saturday"
]

function DetermineDayOfWeek(year, month, day)
{
    let monthAdjustments = [ 0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4 ];
    year -= (month < 3) ? 1 : 0;
    let leapYears = Math.floor(year / 4);
    let gregorianYears = Math.floor(year / 100);
    let modernYears = Math.floor(year / 400);

    return DaysOfWeek[(year + leapYears - gregorianYears + modernYears + monthAdjustments[month - 1] + day) % 7];
}

Answer 4

If I were to provide guidance, I would suggest following this code snippet and proceeding towards the desired date:

let DaysOfWeek=[
    "Sunday",
    "Monday",
    "Tuesday",
    "Wednesday",
    "Thursday",
    "Friday",
    "Saturday"
]

function DetermineDayOfWeek(year, month, day)
{
    let monthAdjustments = [ 0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4 ];
    year -= (month < 3) ? 1 : 0;
    let leapYears = Math.floor(year / 4);
    let gregorianYears = Math.floor(year / 100);
    let modernYears = Math.floor(year / 400);

    return DaysOfWeek[(year + leapYears - gregorianYears + modernYears + monthAdjustments[month - 1] + day) % 7];
}

Answer 5

Answer №3

This code was quickly put together to make it work, but there is room for optimization.

Feel free to check out the DEMO https://example.com/demo-link

The function aims to find the date of the same weekday next year by adding or subtracting days accordingly to get the closest match.

function findNextYearWeekday(currentDate)
{
  var weekDay = currentDate.getDay();
  var tempDate = new Date();
  tempDate.setFullYear(currentDate.getFullYear() + 1);
  
  var previousDays = 0;
  var nextDays = 0;

  // Check Previous Days
  for (var i = 0; i < 7; i++)
  {
    tempDate.setDate(currentDate.getDate() - i);
    if (weekDay == tempDate.getDay())
    {
      previousDays = i;
    }
  }

  // Reset the date
  tempDate = new Date();
  tempDate.setFullYear(currentDate.getFullYear() + 1);

  // Check Next Days
  for (var i = 0; i < 7; i++)
  {
    tempDate.setDate(currentDate.getDate() + i);
    if (weekDay == tempDate.getDay())
    {
      nextDays = i;
    }
  }

  // Reset the date
  tempDate = new Date();
  tempDate.setFullYear(currentDate.getFullYear() + 1);

  // Determine which date is closer
  if (previousDays < nextDays)
    tempDate.setDate(tempDate.getDate() - previousDays);
  else
    tempDate.setDate(tempDate.getDate() + nextDays);

  return tempDate;
}

Answer 6

This code was quickly put together to make it work, but there is room for optimization.

Feel free to check out the DEMO https://example.com/demo-link

The function aims to find the date of the same weekday next year by adding or subtracting days accordingly to get the closest match.

function findNextYearWeekday(currentDate)
{
  var weekDay = currentDate.getDay();
  var tempDate = new Date();
  tempDate.setFullYear(currentDate.getFullYear() + 1);
  
  var previousDays = 0;
  var nextDays = 0;

  // Check Previous Days
  for (var i = 0; i < 7; i++)
  {
    tempDate.setDate(currentDate.getDate() - i);
    if (weekDay == tempDate.getDay())
    {
      previousDays = i;
    }
  }

  // Reset the date
  tempDate = new Date();
  tempDate.setFullYear(currentDate.getFullYear() + 1);

  // Check Next Days
  for (var i = 0; i < 7; i++)
  {
    tempDate.setDate(currentDate.getDate() + i);
    if (weekDay == tempDate.getDay())
    {
      nextDays = i;
    }
  }

  // Reset the date
  tempDate = new Date();
  tempDate.setFullYear(currentDate.getFullYear() + 1);

  // Determine which date is closer
  if (previousDays < nextDays)
    tempDate.setDate(tempDate.getDate() - previousDays);
  else
    tempDate.setDate(tempDate.getDate() + nextDays);

  return tempDate;
}

Using JavaScript to find the weekday of the same date next year

Answer №1

Answer №2

Answer №3

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