How to Get Difference Between Two Dates in Year/Month/Week/Day

How to get difference between two dates in Year/Month/Week/Day?

This is actually quite tricky. A different total number of days can result in the same result. For example:

• 19th June 2008 to 19th June 2010 = 2 years, but also 365 * 2 days

• 19th June 2006 to 19th June 2008 = 2 years, but also 365 + 366 days due to leap years

You may well want to subtract years until you get to the point where you've got two dates which are less than a year apart. Then subtract months until you get to the point where you've got two dates which are less than a month apart.

Further confusion: subtracting (or adding) months is tricky when you might start with a date of "30th March" - what's a month earlier than that?

Even further confusion (may not be relevant): even a day isn't always 24 hours. Daylight saving anyone?

Even further confusion (almost certainly not relevant): even a minute isn't always 60 seconds. Leap seconds are highly confusing...

I don't have the time to work out the exact right way of doing this right now - this answer is mostly to raise the fact that it's not nearly as simple as it might sound.

EDIT: Unfortunately I'm not going to have enough time to answer this fully. I would suggest you start off by defining a struct representing a Period:

public struct Period
{
    private readonly int days;
    public int Days { get { return days; } }
    private readonly int months;
    public int Months { get { return months; } }
    private readonly int years;
    public int Years { get { return years; } }

    public Period(int years, int months, int days)
    {
        this.years = years;
        this.months = months;
        this.days = days;
    }

    public Period WithDays(int newDays)
    {
        return new Period(years, months, newDays);
    }

    public Period WithMonths(int newMonths)
    {
        return new Period(years, newMonths, days);
    }

    public Period WithYears(int newYears)
    {
        return new Period(newYears, months, days);
    }

    public static DateTime operator +(DateTime date, Period period)
    {
        // TODO: Implement this!
    }

    public static Period Difference(DateTime first, DateTime second)
    {
        // TODO: Implement this!
    }
}

I suggest you implement the + operator first, which should inform the Difference method - you should make sure that first + (Period.Difference(first, second)) == second for all first/second values.

Start with writing a whole slew of unit tests - initially "easy" cases, then move on to tricky ones involving leap years. I know the normal approach is to write one test at a time, but I'd personally brainstorm a bunch of them before you start any implementation work.

Allow yourself a day to implement this properly. It's tricky stuff.

Note that I've omitted weeks here - that value at least is easy, because it's always 7 days. So given a (positive) period, you'd have:

int years = period.Years;
int months = period.Months;
int weeks = period.Days / 7;
int daysWithinWeek = period.Days % 7;

(I suggest you avoid even thinking about negative periods - make sure everything is positive, all the time.)

Calculate difference between two dates in form `X years, Y months, Z week, A day`

I don't believe there's anything built into .NET itself which does this in a useful way. TimeSpan (which you'd get from date2 - date1) doesn't have the concept of months etc - it's just a duration in ticks, effectively.

You can use Noda Time for this, but it's not quite finished yet. The period calcuation part may well change further... I'm not sure yet. Let me know if you'd like a Noda Time code sample.

Also be aware that arithmetic using dates is fundamentally tricky. Sometimes it's hard to work out what the right answer should even be...

C# Date Difference in Year, Month, Week and Days

there is a tricky note about relationship of week and month , basically we think than every 4 week is a month but that is not true, depending on month's length , it might be a few days more than 4 weeks(28 days). you need to correct special conditions for months that fall into this problem.

Get the difference between dates in terms of weeks, months, quarters, and years

what about this:

# get difference between dates `"01.12.2013"` and `"31.12.2013"`

# weeks
difftime(strptime("26.03.2014", format = "%d.%m.%Y"),
strptime("14.01.2013", format = "%d.%m.%Y"),units="weeks")
Time difference of 62.28571 weeks

# months
(as.yearmon(strptime("26.03.2014", format = "%d.%m.%Y"))-
as.yearmon(strptime("14.01.2013", format = "%d.%m.%Y")))*12
[1] 14

# quarters
(as.yearqtr(strptime("26.03.2014", format = "%d.%m.%Y"))-
as.yearqtr(strptime("14.01.2013", format = "%d.%m.%Y")))*4
[1] 4

# years
year(strptime("26.03.2014", format = "%d.%m.%Y"))-
year(strptime("14.01.2013", format = "%d.%m.%Y"))
[1] 1

as.yearmon() and as.yearqtr() are in package zoo. year() is in package lubridate.
What do you think?

How to calculate the difference between two dates using PHP?

Use this for legacy code (PHP < 5.3). For up to date solution see jurka's answer below

You can use strtotime() to convert two dates to unix time and then calculate the number of seconds between them. From this it's rather easy to calculate different time periods.

$date1 = "2007-03-24";
$date2 = "2009-06-26";

$diff = abs(strtotime($date2) - strtotime($date1));

$years = floor($diff / (365*60*60*24));
$months = floor(($diff - $years * 365*60*60*24) / (30*60*60*24));
$days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24));

printf("%d years, %d months, %d days\n", $years, $months, $days);

Edit: Obviously the preferred way of doing this is like described by jurka below. My code is generally only recommended if you don't have PHP 5.3 or better.

Several people in the comments have pointed out that the code above is only an approximation. I still believe that for most purposes that's fine, since the usage of a range is more to provide a sense of how much time has passed or remains rather than to provide precision - if you want to do that, just output the date.

Despite all that, I've decided to address the complaints. If you truly need an exact range but haven't got access to PHP 5.3, use the code below (it should work in PHP 4 as well). This is a direct port of the code that PHP uses internally to calculate ranges, with the exception that it doesn't take daylight savings time into account. That means that it's off by an hour at most, but except for that it should be correct.

<?php

/**
 * Calculate differences between two dates with precise semantics. Based on PHPs DateTime::diff()
 * implementation by Derick Rethans. Ported to PHP by Emil H, 2011-05-02. No rights reserved.
 * 
 * See here for original code:
 * http://svn.php.net/viewvc/php/php-src/trunk/ext/date/lib/tm2unixtime.c?revision=302890&view=markup
 * http://svn.php.net/viewvc/php/php-src/trunk/ext/date/lib/interval.c?revision=298973&view=markup
 */

function _date_range_limit($start, $end, $adj, $a, $b, $result)
{
    if ($result[$a] < $start) {
        $result[$b] -= intval(($start - $result[$a] - 1) / $adj) + 1;
        $result[$a] += $adj * intval(($start - $result[$a] - 1) / $adj + 1);
    }

    if ($result[$a] >= $end) {
        $result[$b] += intval($result[$a] / $adj);
        $result[$a] -= $adj * intval($result[$a] / $adj);
    }

    return $result;
}

function _date_range_limit_days($base, $result)
{
    $days_in_month_leap = array(31, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
    $days_in_month = array(31, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);

    _date_range_limit(1, 13, 12, "m", "y", &$base);

    $year = $base["y"];
    $month = $base["m"];

    if (!$result["invert"]) {
        while ($result["d"] < 0) {
            $month--;
            if ($month < 1) {
                $month += 12;
                $year--;
            }

            $leapyear = $year % 400 == 0 || ($year % 100 != 0 && $year % 4 == 0);
            $days = $leapyear ? $days_in_month_leap[$month] : $days_in_month[$month];

            $result["d"] += $days;
            $result["m"]--;
        }
    } else {
        while ($result["d"] < 0) {
            $leapyear = $year % 400 == 0 || ($year % 100 != 0 && $year % 4 == 0);
            $days = $leapyear ? $days_in_month_leap[$month] : $days_in_month[$month];

            $result["d"] += $days;
            $result["m"]--;

            $month++;
            if ($month > 12) {
                $month -= 12;
                $year++;
            }
        }
    }

    return $result;
}

function _date_normalize($base, $result)
{
    $result = _date_range_limit(0, 60, 60, "s", "i", $result);
    $result = _date_range_limit(0, 60, 60, "i", "h", $result);
    $result = _date_range_limit(0, 24, 24, "h", "d", $result);
    $result = _date_range_limit(0, 12, 12, "m", "y", $result);

    $result = _date_range_limit_days(&$base, &$result);

    $result = _date_range_limit(0, 12, 12, "m", "y", $result);

    return $result;
}

/**
 * Accepts two unix timestamps.
 */
function _date_diff($one, $two)
{
    $invert = false;
    if ($one > $two) {
        list($one, $two) = array($two, $one);
        $invert = true;
    }

    $key = array("y", "m", "d", "h", "i", "s");
    $a = array_combine($key, array_map("intval", explode(" ", date("Y m d H i s", $one))));
    $b = array_combine($key, array_map("intval", explode(" ", date("Y m d H i s", $two))));

    $result = array();
    $result["y"] = $b["y"] - $a["y"];
    $result["m"] = $b["m"] - $a["m"];
    $result["d"] = $b["d"] - $a["d"];
    $result["h"] = $b["h"] - $a["h"];
    $result["i"] = $b["i"] - $a["i"];
    $result["s"] = $b["s"] - $a["s"];
    $result["invert"] = $invert ? 1 : 0;
    $result["days"] = intval(abs(($one - $two)/86400));

    if ($invert) {
        _date_normalize(&$a, &$result);
    } else {
        _date_normalize(&$b, &$result);
    }

    return $result;
}

$date = "1986-11-10 19:37:22";

print_r(_date_diff(strtotime($date), time()));
print_r(_date_diff(time(), strtotime($date)));

How to get difference of days/months/years (datediff) between two dates?

SELECT
  AGE('2012-03-05', '2010-04-01'),
  DATE_PART('year', AGE('2012-03-05', '2010-04-01')) AS years,
  DATE_PART('month', AGE('2012-03-05', '2010-04-01')) AS months,
  DATE_PART('day', AGE('2012-03-05', '2010-04-01')) AS days;

This will give you full years, month, days ... between two dates:

          age          | years | months | days
-----------------------+-------+--------+------
 1 year 11 mons 4 days |     1 |     11 |    4

More detailed datediff information.

Get the year, month week, day, hour, minute, second interval on two dates in PHP

You are use PHP date_diff() Function

$date1=date_create("2013-03-15");
$date2=date_create("2013-12-12");
$diff=date_diff($date1,$date2);

Getting the difference between two Dates (months/days/hours/minutes/seconds) in Swift

Xcode 8.3 • Swift 3.1 or later

You can use Calendar to help you create an extension to do your date calculations as follow:

extension Date {
    /// Returns the amount of years from another date
    func years(from date: Date) -> Int {
        return Calendar.current.dateComponents([.year], from: date, to: self).year ?? 0
    }
    /// Returns the amount of months from another date
    func months(from date: Date) -> Int {
        return Calendar.current.dateComponents([.month], from: date, to: self).month ?? 0
    }
    /// Returns the amount of weeks from another date
    func weeks(from date: Date) -> Int {
        return Calendar.current.dateComponents([.weekOfMonth], from: date, to: self).weekOfMonth ?? 0
    }
    /// Returns the amount of days from another date
    func days(from date: Date) -> Int {
        return Calendar.current.dateComponents([.day], from: date, to: self).day ?? 0
    }
    /// Returns the amount of hours from another date
    func hours(from date: Date) -> Int {
        return Calendar.current.dateComponents([.hour], from: date, to: self).hour ?? 0
    }
    /// Returns the amount of minutes from another date
    func minutes(from date: Date) -> Int {
        return Calendar.current.dateComponents([.minute], from: date, to: self).minute ?? 0
    }
    /// Returns the amount of seconds from another date
    func seconds(from date: Date) -> Int {
        return Calendar.current.dateComponents([.second], from: date, to: self).second ?? 0
    }
    /// Returns the a custom time interval description from another date
    func offset(from date: Date) -> String {
        if years(from: date)   > 0 { return "\(years(from: date))y"   }
        if months(from: date)  > 0 { return "\(months(from: date))M"  }
        if weeks(from: date)   > 0 { return "\(weeks(from: date))w"   }
        if days(from: date)    > 0 { return "\(days(from: date))d"    }
        if hours(from: date)   > 0 { return "\(hours(from: date))h"   }
        if minutes(from: date) > 0 { return "\(minutes(from: date))m" }
        if seconds(from: date) > 0 { return "\(seconds(from: date))s" }
        return ""
    }
}

---

Using Date Components Formatter

let dateComponentsFormatter = DateComponentsFormatter()
dateComponentsFormatter.allowedUnits = [.second, .minute, .hour, .day, .weekOfMonth, .month, .year]
dateComponentsFormatter.maximumUnitCount = 1
dateComponentsFormatter.unitsStyle = .full
dateComponentsFormatter.string(from: Date(), to: Date(timeIntervalSinceNow: 4000000))  // "1 month"

---

let date1 = DateComponents(calendar: .current, year: 2014, month: 11, day: 28, hour: 5, minute: 9).date!
let date2 = DateComponents(calendar: .current, year: 2015, month: 8, day: 28, hour: 5, minute: 9).date!

let years = date2.years(from: date1)     // 0
let months = date2.months(from: date1)   // 9
let weeks = date2.weeks(from: date1)     // 39
let days = date2.days(from: date1)       // 273
let hours = date2.hours(from: date1)     // 6,553
let minutes = date2.minutes(from: date1) // 393,180
let seconds = date2.seconds(from: date1) // 23,590,800

let timeOffset = date2.offset(from: date1) // "9M"

let date3 = DateComponents(calendar: .current, year: 2014, month: 11, day: 28, hour: 5, minute: 9).date!
let date4 = DateComponents(calendar: .current, year: 2015, month: 11, day: 28, hour: 5, minute: 9).date!

let timeOffset2 = date4.offset(from: date3) // "1y"

let date5 = DateComponents(calendar: .current, year: 2017, month: 4, day: 28).date!
let now = Date()
let timeOffset3 = now.offset(from: date5) // "1w"

---

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