I use this code:
$today_date = new DateTime('now');
$final_date = new DateTime('2012-03-22 09:00');
$interval = $today_date->diff($final_date);
$time_left = $interval->format('%d:%H:%i:%s');
I want $time_left to have the format dd:hh:mm:ss format (d-> day, h-> hour, m-> minute, s-> seconds).
But when i used %d:%H:%i:%s as format, it sometimes displays single number, that is without 0 padding for minute. For example: 11:14:7:45. Expected result: 11:14:07:45
Any mistakes ?
More a comment: Don't confuse the format of DateInterval with format of DateTime: i is Minutes, numeric, e.g. 1, 3, 59. In case you run into a problem or you use a function you don't know in and out, please consult the manual first.
Per DateInterval, %i is "Minutes, numeric: 1, 3, 59". Try %I which uses leading zeroes.
Related
I've created a timing system for a charity race. I'm trying to find the difference between the start time and the finishers time using PHP. I'm not sure I'm recording the times correctly, but this is the start time i just recorded...
20180808180653
And this is a finisher time...
20180808180654
The difference between them is roughly 1 hour 24, but when i use...
date('h:i:s', $finshTime-$startTime)
I get 03:24:20 not 01:34:20.
Can someone please help?
The date method accepts as "integer Unix timestamp". You are supplying instead a number of seconds (1 in your example).
$start = '20180808180653';
$end = '20180808180654';
$diff = $end - $start;
var_dump($diff); //1
$d = date('h:i:s', $$diff);
var_dump($d); //04:00:01
//the above is wrong. You need to try something like the code below
$dStart = new DateTime($start);
$dEnd = new DateTime($end);
$interval = $dStart->diff($dEnd);
var_dump($interval->format('%h:%i:%s'));
I'd be leery using a string representation of a datetime that looks like that. Convert the whole thing into a date format that makes sense like yyyy-mm-dd hh:mm:ss, or a valid unix time stamp.
Your first approach isn't that far off, you just need to use a strtotime function. I'd guarantee that you can first make an accurate Date or Unix time representation of those strings you are using. Rest should fall into place.
First check if the type of $finshTime and $startTime are integer.
you can use get variable type:
gettype($startTime);
if this is the case try this with ():
$diff_date = date('h:i:s', ($finshTime - $startTime) );
if $startTime and $finshTime are string try this:
$diff_date = date('h:i:s', (strtotime($finshTime) - strtotime($startTime)) );
Im trying to get the difference between 2 differente dates in minutes, but is not outputting correctly.
Ex:
$then = "2017-01-23 18:21:24";
//Convert it into a timestamp.
$then = strtotime($then);
//Get the current timestamp.
$now = time();
//Calculate the difference.
$difference = $now - $then;
//Convert seconds into minutes.
$minutes = floor($difference / 60);
echo $minutes;
Is outputting 611 minutes, and is wrong since from "2017-01-23 18:21:24" to "2017-01-24 12:36:24" it past much more than 611 minutes. Is my code incorrect?
Try to set your default timezone
date_default_timezone_set('Europe/Copenhagen');
Ofc change Europe/Copenhagen for the one that suits your needs.
If you are using or able to use PHP 5.3.x or later, you can use its DateTime object functionality:
$date_a = new DateTime('2010-10-20 08:10:00');
$date_b = new DateTime('2008-12-13 10:42:00');
$interval = date_diff($date_a,$date_b);
echo $interval->format('%h:%i:%s');
You can play with the format in a variety of ways, and once you have dates in DateTime objects, you can take advantage of a lot of different functionality, for example comparison via normal operators. See the manual for more: http://us3.php.net/manual/en/datetime.diff.php
I've checked your code it works perfectly So if have any doubt see your result
But you got wrong, so to ignore this set your timezone.
I've got a problem with my "DateTime difference code":
$timeStart = new DateTime('2015-11-28');
$timeEnd = new DateTime('2016-11-28');
$interval = $timeEnd->diff($timeStart);
$result = $interval->format('%d');
echo $result." day(s)";
When I visualize $result, PHP show me 0. But between those two dates there are more days than 0 day...
php does not calculate the difference between two dates that are not in the same year?
Because there are 0 days difference. There is however a 1 year difference. If you changed %d to %y you'd get 1. So there's a difference of 1 year, 0 months and 0 days.
What you can use instead is the days property on DateInterval, as such:
$result = $interval->days;
Okay, I'm aware the answer was given already. But below is just a bit explanation.
In fact, DateInterval::format() does makes sense when you have a fixed amount of time (in years, months, days, hours), like this:
$interval = new DateInterval('P2Y4DT6H8M');
echo $interval->format('%d days');
That isn't your case!
where you have a relative time (2016-11-28 related to 2015-11-28) at all. In this specific case you want the days amount past since 28-11-2015.
That's why DateInterval::days (DateTime::diff() returns a DateInterval object) makes sense:
$start = new DateTime('2015-11-28');
$end = new DateTime('2016-12-28');
var_dump($end->diff($start)->days);
I want to add an x number of week days (e.g. 48 weekday hours) to the current timestamp. I am trying to do this using the following
echo (strtotime('2 weekdays');
However, this doesn't seem to take me an exact 48 hours ahead in time. For example, inputting the current server time of Tuesday 18/03/2014 10:47 returns Thursday 20/03/2014 00:00. using the following function:
echo (strtotime('2 weekdays')-mktime())/86400;
It can tell that it's returning only 1.3 weekdays from now.
Why is it doing this? Are there any existing functions which allow an exact amount of weekday hours?
Given you want to preserve the weekdays functionality and not loose the hours, minutes and seconds, you could do this:
$now = new DateTime();
$hms = new DateInterval(
'PT'.$now->format('H').'H'.
$now->format('i').'M'.
$now->format('s').'S'.
);
$date = new DateTime('2 weekdays');
$date->add($hms);//add hours here again
The reason why weekday doesn't add the hours is because, if you add 1 weekday at any point in time on a monday, the next weekday has to be tuesday.
The hour simply does not matter. Say your date is 2014-01-02 12:12:12, and you want the next weekday, that day starts at 2014-01-03 00:00:00, so that's what you get.
My last solution works though, and here's how: I use the $now instance of DateTime, and its format method to construct a DateInterval format string, to be passed to the constructor. An interval format is quite easy: it starts with P, for period, then a digit and a char to indicate what that digit represents: 1Y for 1 Year, and 2D for 2 Days.
However, we're only interested in hours, minutes and seconds. Actual time, which is indicated using a T in the interval format string, hence we start the string with PT (Period Time).
Using the format specifiers H, i and s, we construct an interval format that in the case of 12:12:12 looks like this:
$hms = new DateInterval(
'PT12H12M12S'
);
Then, it's a simple matter of calling the DateTime::add method to add the hours, minutes and seconds to our date + weekdays:
$weekdays = new DateTime('6 weekdays');
$weekdays->add($hms);
echo $weekdays->format('Y-m-d H:i:s'), PHP_EOL;
And you're there.
Alternatively, you could just use the basic same trick to compute the actual day-difference between your initial date, and that date + x weekdays, and then add that diff to your initial date. It's the same basic principle, but instead of having to create a format like PTXHXMXS, a simple PXD will do.
Working example here
I'd urge you to use the DateInterface classes, as it is more flexible, allows for type-hinting to be used and makes dealing with dates just a whole lot easier for all of us. Besides, it's not too different from your current code:
$today = new DateTime;
$tomorrow = new DateTime('tomorrow');
$dayAfter = new DateTime('2 days');
In fact, it's a lot easier if you want to do frequent date manipulations on a single date:
$date = new DateTime();//or DateTime::createFromFormat('Y-m-d H:i:s', $dateString);
$diff = new DateInterval('P2D');//2 days
$date->add($diff);
echo $date->format('Y-m-d H:i:s'), PHP_EOL, 'is the date + 2 days', PHP_EOL;
$date->sub($diff);
echo $date->format('Y-m-d H:i:s'), PHP_EOL, 'was the original date, now restored';
Easy, once you've spent some time browsing through the docs
I think I have found a solution. It's primitive but after some quick testing it seems to work.
The function calculates the time passed since midnight of the current day, and adds it onto the date returned by strtotime. Since this could fall into a weekend day, I've checked and added an extra day or two accordingly.
function weekDays($days) {
$tstamp = (strtotime($days.' weekdays') + (time() - strtotime("today")));
if(date('D',$tstamp) == 'Sat') {
$tstamp = $tstamp + 86400*2;
}
elseif(date('D',$tstamp) == 'Sun') {
$tstamp = $tstamp + 86400;
}
return $tstamp;
}
Function strtotime('2 weekdays') seems to add 2 weekdays to the current date without the time.
If you want to add 48 hours why not adding 2*24*60*60 to mktime()?
echo(date('Y-m-d', mktime()+2*24*60*60));
The currently accepted solution works, but it will fail when you want to add weekdays to a timestamp that is not now. Here's a simpler snippet that will work for any given point in time:
$start = new DateTime('2021-09-29 15:12:10');
$start->add(date_interval_create_from_date_string('+ 3 weekdays'));
echo $start->format('Y-m-d H:i:s'); // 2021-10-04 15:12:10
Note that this will also work for a negative amount of weekdays:
$start = new DateTime('2021-09-29 15:12:10');
$start->add(date_interval_create_from_date_string('- 3 weekdays'));
echo $start->format('Y-m-d H:i:s'); // 2021-09-24 15:12:10
$bought_months = 6;
$currentDate = date('Y-m-d');
$expiring = strtotime('+'.$bought_months.' month', $currentDate);
This is what i tried.
I am trying to get a unix timestamp value of 6 months ahead from today.
How can i add months to my unix timestamp right? I tried above, since my other thought - calculating by seconds like one month in seconds: 2 629 743.83 * how many months + current timestamp, will not be precise.
(ofcourse because months have different number of days)
I get "A non well formed numeric value encountered" for the code above.
How can i do this?
Since you are using the current timestamp, you can omit the $currentDate variable altogether and it should make the notice go away too.
$bought_months = 6;
$expiring = strtotime('+'.$bought_months.' month');
To make the date correctly you'd need to use something like:
mktime( 0, 0, 0, ( $month + 6 ), $day, $year );
For maximum compatibility with post-2038 dates, use php's builtin DateTime class:
$d = new DateTime();
$d->modify("+6 months");
echo $d->format("U");