so, I substract two dates:
$d = $start->diff($end);
and now I want to get the date in seconds (not the second parameter). I know it can be done with a native Php function - but it only works above 1970. I know I have to somehow operate with format() method, but I dont get it...
You could access the public properties of the DateInterval class and do some math on it:
$seconds = $d->s + ($d->i * 60) + ($d->h * 3600) + ($d->d * 86400) + ($d->m * 2592000); // and so on
but once we get into the months there will be variance by +/- 2 days unless we stick to an arbitrary definition of a month as 2592000 secs (30 days).
You can also use the difference of the two UNIX timestamps from the DateTime objects (but you will end up having problems with the dates being less than year of 1970):
$seconds = $end->getTimestamp() - $start->getTimestamp();
The following seems to work, although I feel there should be a better way.
$dateIntrvl = $start->diff($end);
$days = $dateIntrvl->format('%a');
$hours = $dateIntrvl->format('%h');
$mins = $dateIntrvl->format('%i');
$secs = $dateIntrvl->format('%s');
$totalSeconds = ($days * 24 * 60 * 60) + ($hours * 60 * 60) + ($mins * 60) + $secs;
And yes, it seems to be working for dates earlier than 1970.
How about using a class DateTime / DateTimeImmutable to create current DateTime instance, then add your DateInterval $d by using the add() method, and then using the getTimestamp() method.
(new \DateTimeImmutable())->add($d)->getTimestamp()
To get number of seconds from now, just subtract time().
(new \DateTimeImmutable())->add($d)->getTimestamp() - \time()
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 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;
}
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);
}
How would i convert a format like:
13 hours and 4 mins ago
Into something i can use in php to further convert?
try
strtotime('13 hours 4 mins ago', time())
http://php.net/manual/en/function.strtotime.php
http://www.php.net/manual/en/datetime.formats.php
http://www.php.net/manual/en/datetime.formats.relative.php
To get a unix timestamp fromt he current server time:
$timestamp = strtotime("-13 hours 4 minutes");
Try the DateInterval class - http://www.php.net/manual/en/class.dateinterval.php - which you can then sub() from a regular DateTime object.
Do it the hard way:
$offset = (time() - ((60 * 60) * 13) + (4 * 60));
Working Example: http://codepad.org/25CJPL76
Explanation:
(60 * 60) is 1 hour, then * 13 to make 13 hours, plus (4 * 60) for the 4 minutes, minus the current time.
What i usually do is create several constants to define the values of specific time values like so:
define("NOW",time());
define("ONE_MIN", 60);
define("ONE_HOUR",ONE_MIN * 60);
define("ONE_DAY", ONE_HOUR * 24);
define("ONE_YEAR",ONE_DAY * 365);
and then do something like:
$offset = (NOW - ((ONE_HOUR * 13) + (ONCE_MIN * 4)));
in regards to actually parsing the string, you should look at a javascript function and convert to PHP, the source is available from PHP.JS:
http://phpjs.org/functions/strtotime:554
Use strtotime(), and perhaps date() later, e.g.:
$thetime = strtotime('13 hours 4 minutes ago');
$thedate = date('Y-m-d', $thetime);
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");