$date1 = "01:45:54";
$date2 = "01:16:07";
$diff = abs(strtotime($date2) + strtotime($date1));
$hours = floor(($diff / (60))/ (60));
$minuts = floor(($diff - $hours*60*60)/ 60);
$seconds = floor(($diff - $hours*60*60 - $minuts*60));
printf("%d hours, %d minuts\n, %d seconds\n", $hours, $minuts, $seconds);
result 885685 hours, 2 minuts , 1 seconds (wrong)
i want result 3 hours, 2 minuts , 1 seconds
Thank you in Advance
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
i have time data in database, i'm recording time with time() function (value 1551866538) on database, and i want write php remaining time for 1 day but as 24 hour. I try date_diff() function but i get error every time, thanks for help.
$mysqltime = $data->registerdate; // 1551866538
$now = time();
$remainingtime = $now - $mysqltime; // I want show as hour
You can also use The DateTime class as below:
$today = new DateTime('now');
$tomorrow = new DateTime(date('Y-m-d H:i:s', '1551866538'));
$difference = $today->diff($tomorrow);
echo $difference->format('%h hours %i minutes remaining');
Output
1 hours 42 minutes remaining
<?php
// Declare and define two dates
$date1 = strtotime("2016-06-01 22:45:00");
$date2 = strtotime("2018-09-21 10:44:01");
// Formulate the Difference between two dates
$diff = abs($date2 - $date1);
// To get the year divide the resultant date into
// total seconds in a year (365*60*60*24)
$years = floor($diff / (365*60*60*24));
// To get the month, subtract it with years and
// divide the resultant date into
// total seconds in a month (30*60*60*24)
$months = floor(($diff - $years * 365*60*60*24)
/ (30*60*60*24));
// To get the day, subtract it with years and
// months and divide the resultant date into
// total seconds in a days (60*60*24)
$days = floor(($diff - $years * 365*60*60*24 -
$months*30*60*60*24)/ (60*60*24));
// To get the hour, subtract it with years,
// months & seconds and divide the resultant
// date into total seconds in a hours (60*60)
$hours = floor(($diff - $years * 365*60*60*24
- $months*30*60*60*24 - $days*60*60*24)
/ (60*60));
// To get the minutes, subtract it with years,
// months, seconds and hours and divide the
// resultant date into total seconds i.e. 60
$minutes = floor(($diff - $years * 365*60*60*24
- $months*30*60*60*24 - $days*60*60*24
- $hours*60*60)/ 60);
// To get the minutes, subtract it with years,
// months, seconds, hours and minutes
$seconds = floor(($diff - $years * 365*60*60*24
- $months*30*60*60*24 - $days*60*60*24
- $hours*60*60 - $minutes*60));
// Print the result
printf("%d years, %d months, %d days, %d hours, "
. "%d minutes, %d seconds", $years, $months,
$days, $hours, $minutes, $seconds);
?>
Output:
2 years, 3 months, 21 days, 11 hours, 59 minutes, 1 seconds
Reference
$mysqltime = $data->registerdate; // 1551866538
$now = time();
$remainingtime = $now - $mysqltime;
echo $hours = round($remainingtime / (60 * 60)); // hours
I have two times, like this:
$date1 = strtotime("02/12/2019 10:10:54 am");
$date2 = strtotime("02/12/2019 10:11:07 pm");
$diff = abs($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));
$hours = floor(($diff - $years * 365*60*60*24
- $months*30*60*60*24 - $days*60*60*24)
/ (60*60));
$minutes = floor(($diff - $years * 365*60*60*24
- $months*30*60*60*24 - $days*60*60*24
- $hours*60*60)/ 60);
$seconds = floor(($diff - $years * 365*60*60*24
- $months*30*60*60*24 - $days*60*60*24
- $hours*60*60 - $minutes*60));
printf(" %d days, %d hours, "
. "%d minutes, %d seconds",
$days, $hours, $minutes, $seconds);
I want to calculate the difference between them, but it's not working.
This is much easier with the DateTime class, which has a diff method which produces a DateInterval object that can output the difference in whatever format you like:
$date1 = new DateTime("02/12/2019 10:10:54 am");
$date2 = new DateTime("02/12/2019 10:11:07 pm");
$diff = $date1->diff($date2);
echo $diff->format('%y years, %m months, %d days, %h hours, %i minutes and %s seconds');
Output:
0 years, 0 months, 0 days, 12 hours, 0 minutes and 13 seconds
Demo on 3v4l.org
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, $days);
?>
Output:
0 years, 1 months, 2 days
I want to subtract one date to another using php and display result in the days - hours - min - sec format. How i do this using php . I tried this using time stamp but it does not give proper values. please give suggestion
For ex : from date 2012-04-27 19:30:56 to date 2012-04-27 19:37:56
i used this code
if(strtotime($history['datetimestamp']) > strtotime($lasttime)) {
$totalelapsed1 = (strtotime($history['datetimestamp'])-strtotime($lasttime));
if($totalelapsed1 > 60 ) {
$sec = $totalelapsed1%60;
$min = round($totalelapsed1/60 , 0);
$minute = $min + $minute;
$second = $sec + $second;
// echo $min. " min " .$sec." sec";
} else {
//echo "0 min " . $totalelapsed1." sec";
$minute = 0 + $minute;
$second = $totalelapsed1 + $second;
}
} else {
$minute = 0 + $minute;
$second = 0 + $second;
// echo "0 min 0 sec";
}
From how to subtract two dates and times to get difference by VolkerK:
You have to use like this:-
<?php
//$now = new DateTime(); // current date/time
$now = new DateTime("2010-07-28 01:11:50");
$ref = new DateTime("2010-07-30 05:56:40");
$diff = $now->diff($ref);
printf('%d days, %d hours, %d minutes', $diff->d, $diff->h, $diff->i);
prints 2 days, 4 hours, 44 minutes
see http://docs.php.net/datetime.diff
edit: But you could also shift the problem more to the database side, e.g. by storing the expiration date/time in the table and then do a query like
... WHERE key='7gedufgweufg' AND expires<Now()
Many rdbms have reasonable/good support for date/time arithmetic.
Link Url:- how to subtract two dates and times to get difference
http://www.webpronews.com/calculating-the-difference-between-two-dates-using-php-2005-11
I suggest to use DateTime and DateInterval objects.
$date1 = new DateTime("2007-03-24");
$date2 = new DateTime("2009-06-26");
$interval = $date1->diff($date2);
echo "days difference ".$interval->d." days ";
read more php DateTime::diff manual
Try using DateTime:diff().
Examples are provided on that page.
This script resolve my issue
$date1 = date("d-m-Y H:i:s",strtotime($date1));
$date2 = date("d-m-Y H:i:s",strtotime($lasttime));
$diff = abs(strtotime($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));
$hours = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24)/ (60*60)); $minuts = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24 - $hours*60*60)/ 60); $seconds = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24 - $hours*60*60 - $minuts*60));
printf("%d years, %d months, %d days, %d hours, %d minuts\n, %d seconds\n", $years, $months, $days, $hours, $minuts, $seconds);