So I have used this method to get the difference between 2 dates.
$diff = abs(strtotime($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));
Now, lets say that I want to convert the years and months into days. How do I do that?
Using DateTime this is a piece of cake:
$date1 = new DateTime($date1);
$date2 = new DateTime($date2);
$diff = $date1->diff($date2, true);
echo $diff->format('%a') . ' days';
$currentDate = date("d-m-Y");
$date1 = date_create("".$joining_date."");
$date2 = date_create("".$currentDate."");
$diff12 = date_diff($date2, $date1);
$hub_days = $diff12->days;
$months = $diff12->m;
$years = $diff12->y;
Related
I'm facing some problem while calculating the difference between two dates because of Date Format, please help me to fix this issue.
Date 1 - (Format: d/m/Y)
date_default_timezone_set("Asia/Kolkata");
$date1 = date('d/m/Y');
//Output - 20/05/2020
Date 2 - (Format: d/m/Y)
$date2 - 01/27/2020
My Code -
$diff = abs($date2 - $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));
// Print the result
printf("%d years, %d months, %d days", $years, $months, $days);
You don't have to calculate days, month etc manually. There is already DateTime Class available in PHP which you can leverage.
$date1 = DateTime::createFromFormat('d/m/Y', '20/05/2020'); // Use $date1 = new DateTime('NOW'); For Current Time
$date2 = DateTime::createFromFormat('d/m/Y', '25/05/2020');
$interval = $date1->diff($date2);
printf("%d years, %d months, %d days", $interval->y, $interval->m, $interval->d);
Official PHP Documentation: PHP DateTime Class
First you shoud use date format(Format: Y/m/d) and second use strtotime to convert date to seconds because abs function working with numeric values. try following code :
date_default_timezone_set("Asia/Kolkata");
$date1 = strtotime(date('Y/m/d'));
$date2 = strtotime('2020/05/27');
$diff = abs($date2 - $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));
// Print the result
printf("%d years, %d months, %d days", $years, $months, $days);
Output :
0 years, 0 months, 7 days
Hey I am trying to get total months from two dates in php. I have searched about leap year calculations between two dates everywhere on internet but did not find the answer.
If my input is "2019-01-01" to "2019-03-31" then the result i expected is 3 months but result i get is 2 month.
Following is my code .
$date1 = strtotime("2019-01-01");
$date2 = strtotime("2019-02-28");
$diff = abs($date2 - $date1);
$years = floor($diff / (365*60*60*24));
$months = floor(($diff - $years * 365*60*60*24)
/ (30*60*60*24));
printf("%d months",$months);
where am i going wrong
I Understand Your Problem..
Actually the result you are getting is correct,because day ends at night 12pm and you are checking at day time.So until day ends you cant get complete month.If You add one day and check you will get correctly 3 Months
Below I Made Some Changes to Your Code..
$new_date = date('Y-m-d', strtotime('2019-02-28' . ' +1 day'));
$date1 = strtotime("2019-01-01");
$date2 = strtotime($new_date);
$diff = abs($date2 - $date1);
$years = floor($diff / (365*60*60*24));
$months = floor(($diff - $years * 365*60*60*24)
/ (30*60*60*24));
printf("%d months",$months);
You can Also Get it by Mysql querying to server
SELECT TIMESTAMPDIFF(MONTH, '2019-01-01', (SELECT DATE_ADD('2019-02-28', INTERVAL 1 DAY))) as month
You can Try it.Hope it Helps.
Your math calculates the number of months between the two dates, exclusive of the ending month.
Add 1 to the total, like so.
$months = (floor(($diff - $years * 365*60*60*24)
/ (30*60*60*24))) +1;
You can try the following:
$date1 = "2019-01-01";
$date2 = "2019-02-28";
$timestamp1 = strtotime($date1);
$timestamp2 = strtotime($date2);
$year1 = date('Y', $timestamp1);
$year2 = date('Y', $timestamp2);
$month1 = date('m', $timestamp1);
$month2 = date('m', $timestamp2);
$diff = (($year2 - $year1) * 12) + ($month2 - $month1);
printf("%d months",$diff);
Try it here: http://sandbox.onlinephpfunctions.com/code/496a08612489977e9e23e8ef3ea07ba991bc5e23
You have to add extra month in your $month variable
$date1 = strtotime("2019-01-01");
$date2 = strtotime("2019-03-31");
$diff = abs($date2 - $date1);
$years = floor($diff / (365*60*60*24));
$months = floor(($diff - $years * 365*60*60*24)
/ (30*60*60*24)) +1 ;
printf("%d months",$months); //Output : give add one extra
OR
$date1 = date_create('2019-01-01');
$date2 date_create('2019-03-31');
$interval= date_diff($date1, $date1);
echo $interval->format('%m months');
$date2 = $row['returnbefore'];
$date1 = date('Y/m/d');
$diff = abs(strtotime($date1) - strtotime($date2));
$years = floor($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));
$final = ($days * 10) ;
echo $final;
i want to do a library system when user return book will check the return date between current date and calculate a fine to user which multiply day by 10 .
what step i left ? and i wan set the current time is my computer time so i can test it
This is a fairly simple problem to accomplish using DateTime and DateInterval.
$now = new DateTime();
$dueDate = new DateTime($row['returnbefore']);
$lateInterval = $now - $dueDate;
$daysLate = $lateInterval->format('d');
$fine = $daysLate > 0 ? intval(floor($daysLate)) * 10 : 0; // This means that you are not charged for a late day until the end of the day
I tried with the below code to find the difference between two dates which is passed through post variable and print, but failed.
$fromdate=$_POST['from_date'];
$todate=$_POST['to_date'];
$date1 = new DateTime($fromdate); //inclusive
$date2 = new DateTime($todate); //exclusive
$diff = $date2->diff($date1);
echo $diff;
Something like this should work for you:
<?php
$_POST['from_date'] = "2014-10-01";
$_POST['to_date'] = "2014-11-02";
$fromdate = $_POST['from_date'];
$todate = $_POST['to_date'];
$diff = abs(strtotime($fromdate) - strtotime($todate));
$years = floor($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);
?>
Output:
0 years, 1 months, 2 days
Using the PHP code below, I would expect to get '2' as my output. But I get '1'.
Does anyone know why this is?
$returndate = preg_replace('#(\d+)/(\d+)/(\d+)#', '$3-$2-$1', '2011-03-28');
$departdate = preg_replace('#(\d+)/(\d+)/(\d+)#', '$3-$2-$1', '2011-03-26');
$diff = abs(strtotime($returndate) - strtotime($departdate));
$years = floor($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));
echo $days; // expecting 2, but get 1
Many thanks for any help.
So much calculations... I assume its a rounding issue you have, rounding all the time measurements... here is a simpler look on what you are doing:
function dateDiff($start, $end) {
$start_ts = strtotime($start);
$end_ts = strtotime($end);
$diff = $end_ts - $start_ts;
return round($diff / 86400);
}
$d1 = new DateTime('2011-03-28');
$d2 = new DateTime('2011-03-26');
echo $d1->diff($d2)->d;
Output: 2