How to Figure Out the Day Difference in the Following Example

How can I figure out the day difference in the following example

You can use Calendar method date BySettingHour to get the noon time for your startDate and endDate, to make your calendar calculations not time sensitive:

Xcode 8.2.1 • Swift 3.0.2

extension Date {
    var noon: Date? {
        return Calendar.autoupdatingCurrent.date(bySettingHour: 12, minute: 0, second: 0, of: self)
    }
    func daysBetween(_ date: Date) -> Int? {
        guard let noon = noon, let date = date.noon else { return nil }
         return Calendar.autoupdatingCurrent.dateComponents([.day], from: noon, to: date).day
    }
}

let startDateString = "2016-06-10 01:39:07 +0000"
let todayString = "2016-06-11 00:41:41 +0000"
let formatter = DateFormatter()
formatter.calendar = Calendar(identifier: .iso8601)
formatter.locale = Locale(identifier: "ex_US_POSIX")
formatter.dateFormat = "yyyy-MM-dd HH:mm:ss xxxx"
if let startDate = formatter.date(from: startDateString),
    let endDate = formatter.date(from: todayString),
    let days = startDate.daysBetween(endDate) {
    print(startDate)  // "2016-06-10 01:39:07 +0000\n"
    print(endDate)    // "2016-06-11 00:41:41 +0000\n"
    print(days ?? "nil") // 1
}

---

Swift 2.x

extension NSDate {
    var noon: NSDate {
        return NSCalendar.currentCalendar().dateBySettingHour(12, minute: 0, second: 0, ofDate: self, options: [])!
    }
}
func daysBetweenDate(startDate: NSDate, endDate: NSDate) -> Int {
    return NSCalendar.currentCalendar().components([.Day],
                                fromDate: startDate.noon,
                                toDate: endDate.noon,
                                options: []).day
}

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 calculate number of days between two dates

http://momentjs.com/ or https://date-fns.org/

From Moment docs:

var a = moment([2007, 0, 29]);
var b = moment([2007, 0, 28]);
a.diff(b, 'days')   // =1

or to include the start:

a.diff(b, 'days')+1   // =2

Beats messing with timestamps and time zones manually.

Depending on your specific use case, you can either

1. Use a/b.startOf('day') and/or a/b.endOf('day') to force the diff to be inclusive or exclusive at the "ends" (as suggested by @kotpal in the comments).
2. Set third argument true to get a floating point diff which you can then Math.floor, Math.ceil or Math.round as needed.
3. Option 2 can also be accomplished by getting 'seconds' instead of 'days' and then dividing by 24*60*60.

Calculate difference between two dates (number of days)?

Assuming StartDate and EndDate are of type DateTime:

(EndDate - StartDate).TotalDays

How to calculate number of days between two given dates

If you have two date objects, you can just subtract them, which computes a timedelta object.

from datetime import date

d0 = date(2008, 8, 18)
d1 = date(2008, 9, 26)
delta = d1 - d0
print(delta.days)

The relevant section of the docs:
https://docs.python.org/library/datetime.html.

See this answer for another example.

Calculating the difference between two Java date instances

The JDK Date API is horribly broken unfortunately. I recommend using Joda Time library.

Joda Time has a concept of time Interval:

Interval interval = new Interval(oldTime, new Instant());

EDIT: By the way, Joda has two concepts: Interval for representing an interval of time between two time instants (represent time between 8am and 10am), and a Duration that represents a length of time without the actual time boundaries (e.g. represent two hours!)

If you only care about time comparisions, most Date implementations (including the JDK one) implements Comparable interface which allows you to use the Comparable.compareTo()

Calculate days between two Dates in Java 8

If you want logical calendar days, use DAYS.between() method from java.time.temporal.ChronoUnit:

LocalDate dateBefore;
LocalDate dateAfter;
long daysBetween = DAYS.between(dateBefore, dateAfter);

If you want literal 24 hour days, (a duration), you can use the Duration class instead:

LocalDate today = LocalDate.now()
LocalDate yesterday = today.minusDays(1);
// Duration oneDay = Duration.between(today, yesterday); // throws an exception
Duration.between(today.atStartOfDay(), yesterday.atStartOfDay()).toDays() // another option

For more information, refer to this document.

Finding the number of days between two dates

$now = time(); // or your date as well
$your_date = strtotime("2010-01-31");
$datediff = $now - $your_date;

echo round($datediff / (60 * 60 * 24));

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