I am trying to get the PHP "DateInterval" value in "total minutes" value. How to get it? Seems like simple format("%i minutes") not working?
Here is the sample code:
$test = new \DateTime("48 hours");
$interval = $test->diff(new \DateTime());
Now if I try to get the interval in total days, its fine:
echo $interval->format('%a total days');
It is showing 2 days as output, which is totally fine. What I am trying to get if to get the value in "total minutes", so I tried:
echo $interval->format('%i total minutes');
Which is not working. Any help appreciated to get my desired output.
abs((new \DateTime("48 hours"))->getTimestamp() - (new \DateTime)->getTimestamp()) / 60
That's the easiest way to get the difference in minutes between two DateTime instances.
If you are stuck in a position where all you have is the DateInterval, and you (like me) discover that there seems to be no way to get the total minutes, seconds or whatever of the interval, the solution is to create a DateTime at zero time, add the interval to it, and then get the resulting timestamp:
$timeInterval = //the DateInterval you have;
$intervalInSeconds = (new DateTime())->setTimeStamp(0)->add($timeInterval)->getTimeStamp();
$intervalInMinutes = $intervalInSeconds/60; // and so on
I wrote two functions that just calculates the totalTime from a DateInterval.
Accuracy can be increased by considering years and months.
function getTotalMinutes(DateInterval $int){
return ($int->d * 24 * 60) + ($int->h * 60) + $int->i;
}
function getTotalHours(DateInterval $int){
return ($int->d * 24) + $int->h + $int->i / 60;
}
Here is the excepted answer as a method in PHP7.2 style:
public static function getMinutesDifference(\DateTime $a, \DateTime $b): int
{
return abs($a->getTimestamp() - $b->getTimestamp()) / 60;
}
That works perfectly.
function calculateMinutes(DateInterval $int){
$days = $int->format('%a');
return ($days * 24 * 60) + ($int->h * 60) + $int->i;
}
This question is about minutes but if you want to recalculate every carry over points (like I needed to) you can use this solution suggested by #glavic in the comments on the php.net man page (simplified and turned into a function):
private function calculateCarryOverPoints(\DateInterval $dateInterval): \DateInterval
{
$from = new \DateTime;
$to = clone $from;
// Add time of dateInterval to empty DateTime object
$to = $to->add($dateInterval);
// Calculate difference between zero DateTime and DateTime with added DateInterval time
// Which returns a DateInterval object $diff with correct carry over points (days, hours, minutes, seconds etc.)
return $from->diff($to);
}
Related
I need to get different time lengths in minutes from a few timestamps that are:
starttime
endtime
starttime2nd
endtime2nd
Edit: clarification:
Each time is originally stored as a datetime string, like
"2018-02-21T19:45:13+00:00".
I take that and convert it to strtotime("2018-02-21T19:45:13+00:00");
And from that I get a timestamp : 1519242313
It doesn't seem that I can use plus or minus operators to add or subtract timestamps, like:
$length = ($endtime2nd - $starttime2nd) + ($endtime - $starttime)
Am I required to instantiate DateTime and use the "->diff" method to get a time interval?
I could get one time interval by doing this:
$date1 = new DateTime();
$date2 = new DateTime();
$starttime= $date1->setTimestamp($starttime);
$endtime= $date2->setTimestamp($endtime);
$length = $endtime->diff($starttime);
Does this mean that I need to instantiate four DateTimes to get the total length, and set four timestamps, then get the "->diff" for the two time intervals, and then add them using the "->add" method of DateTime method?
I would just like to know if there is a simpler method?
I don't think you need to instantiate DateTime and use the "->diff" method, because you already have timestamp (as i can see you are using setTimestamp)
<?php
$length = ($endtime2nd - $starttime2nd) + ($endtime - $starttime);
echo round(abs($length / 60), 2). " minute";
?>
You can get the total minuts like this.
$min = $length->d * 24 * 60;
$min = $min + $length->h * 60;
$min = $min + $length->m;
I'm trying to calculate the number of hours a person has worked on a given day. To do so, I need to get the difference between two DateTime objects in Hours, Minutes, and Seconds. So far I can successfully do so like this
$timeIn = new DateTime($time['timeIn']);
$timeOut = new DateTime($time['timeOut']);
$time['hours'] = date_diff($timeIn, $timeOut) -> format("%H:%i:%s");
This seems to work fine, until I put in a test case where the employee forgot to clock out. Now, let's say that
$timeIn = '2016-09-28 14:26:17'
$timeOut = '2016-09-30 09:53:53'
In that case, the difference SHOULD be 43:27:36 (Because there is over a day in between the timeIn and timeOut). Instead, I get 19:27:36 (as if it's just truncating the days off and returning the rest). How can I add that day onto the hours instead of truncating it? (I'm looking to get 43:27:36, NOT 1day, 19 hours, ect ect. So I'm trying to get the answer in HH:MM:SS)
As Scott suggested but with a tweak, we'll need to format this ourselves, but we have nifty sprintf to help:
$start = new \DateTime("2016-09-28 14:26:17");
$end = new \DateTime("2016-09-30 09:53:53");
$interval = $end->diff($start);
$time = sprintf(
'%d:%02d:%02d',
($interval->d * 24) + $interval->h,
$interval->i,
$interval->s
);
I prefer using the \DateTime objects - although there shouldn't be much of a difference in the end date_diff is just an alias of \DateTime::diff.
You want to check the number of days in the DateInterval object;
$start = new \DateTime("2016-09-28 14:26:17");
$end = new \DateTime("2016-09-30 09:53:53");
$interval = $end->diff($start);
$days = $interval->d;
if ($days > 0) {
echo $interval->format("%a %h:%i:%s\n");
} else {
echo $interval->format("%h:%i:%s\n");
}
date_diff() returns a DateInterval instance. Look at DateInterval::format method or DateInterval::days field. Hour, minute and second values refer to the remainder in the last day (if more than one). You are looking for a the total number of days.
$timeIn = new DateTime($time['timeIn']);
$timeOut = new DateTime($time['timeOut']);
$diff = date_diff($timeIn, $timeOut) -> format("%H:%i:%s");
$days = $diff->days;
$time['hours'] = // calculate based on days + remainder...
I'm totalling up time like so.
$totalTimeDiff = new DateTime("#0");
foreach($dbrecords as $row)
{
$timeDiff = date_diff( ... two datetimes from my database ... )
$totalTimeDiff->add($timeDiff);
}
So $totalTimeDiff is a DateTime object with the sum of all of the time differences added together (so a sum of all of the durations). How can I get the total time in seconds?
Why not keep it simple?
$totalseconds=0;
foreach($dbrecords as $row)
$totalseconds+=(UNIX_TIMESTAMP(second_datetime)-UNIX_TIMESTAMP(first_datetime));
use strtotime function
echo strtotime('01:00:00') - strtotime('TODAY');
$totalTimeDiff->format('U');
Taking moonwave99's advice, I used DateInterval (can't remember why I went with DateTime for that in the first place, possibly a workaround for something at another stage of the project) and computed the seconds by adding each value to the total after converting it to seconds (converting hours and minutes to seconds and summing them up). I did this by using the DateInterval class's seconds property as well as the following function to convert a DateInterval to seconds (Note: only accounted for days, hours, minutes, and seconds for my specific case as there's no chance the amount will exceed one month):
function convertDateIntervalToSeconds($dateInterval)
{
$days = $dateInterval->d * 24 * 60 * 60;
$hours = $dateInterval->h * 60 * 60;
$minutes = $dateInterval->i * 60;
$seconds = $dateInterval->s;
return $hours + $minutes + $seconds;
}
$datetime1 = date_create('2009-10-11');
$datetime2 = date_create('2009-10-13');
$interval = date_diff($datetime1, $datetime2);
How do i convert the above $interval to seconds in php
Another way to get the number of seconds in an interval is to add it to the zero date, and get the timestamp of that date:
$seconds = date_create('#0')->add($interval)->getTimestamp();
This method will handle intervals created via the DateInterval contructor more or less correctly, whereas shiplu's answer will ignore years, months and days for such intervals. However, shiplu's answer is more accurate for intervals that were created by subtracting two dates. For intervals consisting only of hours, minutes and seconds, both methods will get the correct answer.
There is a function format for this. But it wont return the number of seconds. To get number of seconds you use this technique
$seconds = abs($datetime1->getTimestamp()-$datetime2->getTimestamp());
If you really want to use $interval you need to calculate it.
$seconds = $interval->days*86400 + $interval->h*3600
+ $interval->i*60 + $interval->s;
Here
86400 is the number of seconds in a day
3600 is the number of seconds in an hour
60 is the number of seconds in a minute
I would only add to shiplu's answer:
function dateIntervalToSeconds($interval)
{
$seconds = $interval->days*86400 + $interval->h*3600
+ $interval->i*60 + $interval->s;
return $interval->invert == 1 ? $seconds*(-1) : $seconds;
}
To handle negative intervals.
Note that - contrary to Brilliand's answer - The code above will consider correctly years, months and dates. Because $interval->days is an absolute value ($interval->d is relative to the month).
EDIT: this function is still not correct, as pointed out by #Brilliand. A counter-example is
new DateInterval('P4M3DT2H');
It doesn't handle months well.
2009-10-05 18:11:08
2009-10-05 18:07:13
This should generate 235,how to do it ?
With DateTime objects, you can do it like this:
$date = new DateTime( '2009-10-05 18:07:13' );
$date2 = new DateTime( '2009-10-05 18:11:08' );
$diffInSeconds = $date2->getTimestamp() - $date->getTimestamp();
You can use strtotime() to do that:
$diff = strtotime('2009-10-05 18:11:08') - strtotime('2009-10-05 18:07:13')
A similar approach is possible with DateTime objects, e.g.
$date = new DateTime( '2009-10-05 18:07:13' );
$date2 = new DateTime( '2009-10-05 18:11:08' );
$diff = $date2->getTimestamp() - $date->getTimestamp();
PHP Date Time reference is helpful for things like this: PHP Date Time Functions
strtotime() is probably the best way.
$seconds = strtotime('2009-10-05 18:11:08') - strtotime('2009-10-05 18:07:13')
For those worrying about the limitations of using timestamps (i.e. using dates before 1970 and beyond 2038), you can simply calculate the difference in seconds like so:
$start = new DateTime('2009-10-05 18:11:08');
$end = new DateTime('2009-10-05 18:07:13');
$diff = $end->diff($start);
$daysInSecs = $diff->format('%r%a') * 24 * 60 * 60;
$hoursInSecs = $diff->h * 60 * 60;
$minsInSecs = $diff->i * 60;
$seconds = $daysInSecs + $hoursInSecs + $minsInSecs + $diff->s;
echo $seconds; // output: 235
Wrote a blog post for those interested in reading more.
Because of unix epoch limitations, you could have problems compairing dates before 1970 and after 2038. I choose to loose precision (=don't look at the single second) but avoid to pass trough unix epoch conversions (getTimestamp). It depends on what you are doing to do...
In my case, using 365 instead (12*30) and "30" as mean month lenght, reduced the error in an usable output.
function DateIntervalToSec($start,$end){ // as datetime object returns difference in seconds
$diff = $end->diff($start);
$diff_sec = $diff->format('%r').( // prepend the sign - if negative, change it to R if you want the +, too
($diff->s)+ // seconds (no errors)
(60*($diff->i))+ // minutes (no errors)
(60*60*($diff->h))+ // hours (no errors)
(24*60*60*($diff->d))+ // days (no errors)
(30*24*60*60*($diff->m))+ // months (???)
(365*24*60*60*($diff->y)) // years (???)
);
return $diff_sec;
}
Note that the error could be 0, if "mean" quantities are intended for diff. The PHP docs don't speaks about this...
In a bad case, error could be:
0 seconds if diff is applied to time gaps < 1 month
0 to 3 days if diff is applied to time gaps > 1 month
0 to 14 days if diff is applied to time gaps > 1 year
I prefer to suppose that somebody decided to consider "m" as 30 days and "y" as 365, charging "d" with the difference when "diff" walk trough non-30-days months...
If somebody knows something more about this and can provide official documentation, is welcome!
strtotime("2009-10-05 18:11:08") - strtotime("2009-10-05 18:07:13")
The solution proposed by #designcise is wrong when "end date" is before "start date".
Here is the corrected calculation
$diff = $start->diff($end);
$daysInSecs = $diff->format('%r%a') * 24 * 60 * 60;
$hoursInSecs = $diff->format('%r%h') * 60 * 60;
$minsInSecs = $diff->format('%r%i') * 60;
$seconds = $daysInSecs + $hoursInSecs + $minsInSecs + $diff->format('%r%s');
A simple and exact solution (exemplifying Nilz11's comment):
$hiDate = new DateTime("2310-05-22 08:33:26");
$loDate = new DateTime("1910-11-03 13:00:01");
$diff = $hiDate->diff($loDate);
$secs = ((($diff->format("%a") * 24) + $diff->format("%H")) * 60 +
$diff->format("%i")) * 60 + $diff->format("%s");