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));
Date Difference in php on days?
strtotime will convert your date string to a unix time stamp. (seconds since the unix epoch.
$ts1 = strtotime($date1);
$ts2 = strtotime($date2);
$seconds_diff = $ts2 - $ts1;
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($date)));
difference between two dates is not working in php
You are creating two dates within the same year. A full year has not elapsed. It is TO the end date, not THROUGH the end date.
You need to set your end date to be January 1, 2016 as it is not including the full day of December 31, 2015; it's just evaluating to 2015-12-31 00:00:00.000000.
// add a day to the end date to ensure the final date is included in the comparison
date_add($date2,date_interval_create_from_date_string("1 day"));
PHP date difference calculate onlys for days
PHP:
<?php
$otherday = date_create("01-05-2015");
$today= date_create(date("d-m-Y"));
$days = $today->diff($otherday);
echo $days->format("%R%a");
?>
%R means (+) or (-) you can use only %a.
How to calculate the number of days (difference) between two dates in PHP?
$start = strtotime('2015-03-13');
$end = strtotime('2015-03-20');
$diff = $end - $start;
$days = floor($diff / (3600 * 24));
Calculating MySQL and PHP date difference in days
try this
$daydiff=floor((abs(strtotime(date("Y-m-d")) - strtotime($ArrivalDate))/(60*60*24)));
just change your current date function format so it will give your correct answer means 26 days.
How to get time difference in minutes in PHP
Subtract the past most one from the future most one and divide by 60.
Times are done in Unix format so they're just a big number showing the number of seconds from January 1, 1970, 00:00:00 GMT
How to finding the number of days between two days of the week?
Your task doesn't seem to require date functions at all. A simple lookup array will suffice.
- Subtract the starting day's integer value from the ending day's integer.
- If the difference would be zero or less, add 7 to always return the correct, positive day count.
Code: (Demo)
function daysUntil($start, $end) {
$lookup = [
'Sunday' => 0,
'Monday' => 1,
'Tuesday' => 2,
'Wednesday' => 3,
'Thursday' => 4,
'Friday' => 5,
'Saturday' => 6
];
$days = $lookup[$end] - $lookup[$start] + ($lookup[$end] <= $lookup[$start] ? 7 : 0);
return "{$days} days from {$start} to {$end}\n";
}
echo daysUntil('Wednesday', 'Saturday'); // Thursday, Friday, Saturday
echo daysUntil('Monday', 'Friday'); // Tuesday, Wednesday, Thursday, Friday
echo daysUntil('Thursday', 'Thursday'); // [assumed next week]
echo daysUntil('Friday', 'Monday'); // Saturday, Sunday, Monday
echo daysUntil('Saturday', 'Sunday'); // Sunday
echo daysUntil('Sunday', 'Saturday'); // Monday, Tuesday, Wednesday, Thursday, Friday, Saturday
echo daysUntil('Sunday', 'Wednesday'); // Monday, Tuesday, Wednesday
Output:
3 days from Wednesday to Saturday
4 days from Monday to Friday
7 days from Thursday to Thursday
3 days from Friday to Monday
1 days from Saturday to Sunday
6 days from Sunday to Saturday
3 days from Sunday to Wednesday
Or you can replace the lookup array with 4 function calls and achieve the same outcome: (Demo)
function daysUntil($start, $end) {
$days = date('w', strtotime($end)) - date('w', strtotime($start));
$days += $days < 1 ? 7 : 0;
return "{$days} days from {$start} to {$end}\n";
}
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