Count the days between two dates [duplicate] - php

Question

This question already has answers here:
Finding the number of days between two dates
(34 answers)
Closed 3 months ago.
I want to do is to count the days between two dates excluding the weekends and i;m done doing that using the function below. But whenever the $startDate is greater than the $endDate i can't get the proper result. I try to use if ($startDate>$endDate) and i'm stock with that condition and honestly don't know what is the next step.
function getWorkingDays($startDate,$endDate){
// do strtotime calculations just once
$startDate = strtotime($startDate);
$endDate = strtotime($endDate);
//The total number of days between the two dates. We compute the no. of seconds and divide it to 60*60*24
//We add one to inlude both dates in the interval.
$days = ($endDate - $startDate) / 86400 + 0;
$no_full_weeks = floor($days / 7);
$no_remaining_days = fmod($days, 7);
//It will return 1 if it's Monday,.. ,7 for Sunday
$the_first_day_of_week = date("N", $startDate);
$the_last_day_of_week = date("N", $endDate);
// If one of the value is empty it will return "0"
if ($startDate == '' || $endDate == '')
return "0"; // Default value
//---->The two can be equal in leap years when february has 29 days, the equal sign is added here
//In the first case the whole interval is within a week, in the second case the interval falls in two weeks.
if ($the_first_day_of_week <= $the_last_day_of_week) {
if ($the_first_day_of_week <= 6 && 6 <= $the_last_day_of_week) $no_remaining_days--;
if ($the_first_day_of_week <= 7 && 7 <= $the_last_day_of_week) $no_remaining_days--;
}
else {
// (edit by Tokes to fix an edge case where the start day was a Sunday
// and the end day was NOT a Saturday)
// the day of the week for start is later than the day of the week for end
if ($the_first_day_of_week == 7) {
// if the start date is a Sunday, then we definitely subtract 1 day
$no_remaining_days--;
if ($the_last_day_of_week == 6) {
// if the end date is a Saturday, then we subtract another day
$no_remaining_days--;
}
}
else {
// the start date was a Saturday (or earlier), and the end date was (Mon..Fri)
// so we skip an entire weekend and subtract 2 days
$no_remaining_days -= 2;
}
}
//The no. of business days is: (number of weeks between the two dates) * (5 working days) + the remainder
//---->february in none leap years gave a remainder of 0 but still calculated weekends between first and last day, this is one way to fix it
$workingDays = $no_full_weeks * 5;
if ($no_remaining_days > 0 )
{
$workingDays += $no_remaining_days;
}
return $workingDays;
}

Answer 1

$startTimeStamp = strtotime("2011/07/01");
$endTimeStamp = strtotime("2011/07/17");
$timeDiff = abs($endTimeStamp - $startTimeStamp);
$numberDays = $timeDiff/86400; // 86400 seconds in one day
// and you might want to convert to integer
$numberDays = intval($numberDays);
OR
function dateDiff($start, $end) {
$start_ts = strtotime($start);
$end_ts = strtotime($end);
$diff = $end_ts - $start_ts;
return round($diff / 86400);
}
echo dateDiff("2011-02-15", "2012-01-16").'days';
//Get number of days deference between current date and given date.
echo dateDiff("2011-02-15", date('Y-m-d')).'days';
For Count days excluding use below code
$start = new DateTime('7/17/2017');
$end = new DateTime('7/24/2017');
$oneday = new DateInterval("P1D");
$daysName = array('Mon', 'Tue', 'Wed', 'Thu', 'Fri');
$days = array();
foreach(new DatePeriod($start, $oneday, $end->add($oneday)) as $day) {
$day_num = $day->format("N"); /* 'N' number days 1 (mon) to 7 (sun) */
if($day_num < 6) { /* weekday */
$days[$day->format("Y-m-d")] = date('D', strtotime($day->format("Y-m-d")));;
}
}
echo "<pre>";
print_r($days);
echo count($days);

Answer 2

This will check whether start date is less than end date. If yes, then it will display the days.
<?php
if($days = getWorkingDays("2017-05-01","2018-01-01")){
echo $days;
}
function getWorkingDays($startDate,$endDate){
$startDate = strtotime($startDate);
$endDate = strtotime($endDate);
if($startDate <= $endDate){
$datediff = $endDate - $startDate;
return floor($datediff / (60 * 60 * 24));
}
return false;
}
?>
Output: 245
Functions Used:
strtotime(): The strtotime() function parses an English textual datetime into a Unix timestamp
floor(): The floor() function rounds a number DOWN to the nearest integer
Edit-1: Getting Days After Excluding Weekends (saturdays & sundays)
//getWorkingDays(start_date, end_date)
if($days = getWorkingDays("2017-05-01","2018-01-01")){
echo $days;
}
function getWorkingDays($startDate,$endDate){
$days = false;
$startDate = strtotime($startDate);
$endDate = strtotime($endDate);
if($startDate <= $endDate){
$datediff = $endDate - $startDate;
$days = floor($datediff / (60 * 60 * 24)); // Total Nos Of Days
$sundays = intval($days / 7) + (date('N', $startDate) + $days % 7 >= 7); // Total Nos Of Sundays Between Start Date & End Date
$saturdays = intval($days / 7) + (date('N', $startDate) + $days % 6 >= 6); // Total Nos Of Saturdays Between Start Date & End Date
$days = $days - ($sundays + $saturdays); // Total Nos Of Days Excluding Weekends
}
return $days;
}
?>
Sources:
calculate sundays between two dates
The intval() function is used to get the integer value of a variable.
See Description date('N', $date) : N - The ISO-8601 numeric representation of a day (1 for Monday, 7 for Sunday)

Answer 3

There is the script of code for do this.
<?php
$now = time(); // or your date as well
$your_date = strtotime("2010-01-01");
$datediff = $now - $your_date;
echo floor($datediff / (60 * 60 * 24));
?>
or
$datetime1 = new DateTime("2010-06-20");
$datetime2 = new DateTime("2011-06-22");
$difference = $datetime1->diff($datetime2);
echo 'Difference: '.$difference->y.' years, '
.$difference->m.' months, '
.$difference->d.' days';
print_r($difference);

Answer 4

try this
public function datediff($sdate,$edate){
$diffformat='%a';
$date1 = date_create($sdate);
$date2 = date_create($edate);
$diff12 = date_diff($date2, $date1);
$days = $diff12->format($diffformat) + 1;}

Answer 5

EDIT: I noticed in the comment you want to exclude the weekend day/days (however you didn't mention that in your post !)
you can add the number of days you want to exclude from the week
you can use DateTime::diff and use the option for absolute result (positive difference always)
<?php
function daysBetween2Dates($date1, $date2, $execludedDaysFromWeek = 0)
{
try{
$datetime1 = new \DateTime($date1);
$datetime2 = new \DateTime($date2);
}catch (\Exception $e){
return false;
}
$interval = $datetime1->diff($datetime2,true);
$days = $interval->format('%a');
if($execludedDaysFromWeek < 0 || $execludedDaysFromWeek > 7){
$execludedDaysFromWeek = 0 ;
}
return ceil($days * (7-$execludedDaysFromWeek) / 7);
}
Usage Example
// example 1 : without weekend days, start date is the first one
$days = daysBetween2Dates('2016-12-31','2017-12-31');
echo $days;
// example 2 : without weekend days, start date is the second one
$days = daysBetween2Dates('2017-12-31', '2016-12-31');
echo "<br>\n" .$days;
// example 3 : with weekend days, it returns 6 days for the week
$days = daysBetween2Dates('2017-12-31', '2017-12-24',-1);
echo "<br>\n" .$days;
exit;
this outputs
365
365
6
live demo (https://eval.in/835862)

Answer 6

use date_diff() which returns the difference between two DateTime objects.
$diff=date_diff($startDate,$endDate);