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);
Related
I've not seen anything like this before. This is part of a function that returns the expected answer (a series of five dates)... sometimes. For example, it's been run at 6am, and the result is sometimes incorrect: one of the five dates, either at the first or last, can be missing. Other times, it's fine. Same code, run just a few hours later.
I know working with dates can be a lot more complicated than it first appears, but this has stumped me. I can only hope my inexperience with the DateTime objects is to blame.
$start = new \DateTime(date("Y-m-d", strtotime("-1 day")));
$end = new \DateTime(date("Y-m-d", strtotime("-5 days")));
$diff = $end->diff($start);
$interval = \DateInterval::createFromDateString('-1 day');
$period = new \DatePeriod($start, $interval, $diff->days);
foreach($period as $date) {
echo $date->format("Y-m-d"); // Sometimes first or last date will be missing
}
So for example, if the code is run between 2020-07-05 00:00:00 and 2020-07-05 23:59:59, it should return the last five dates:
2020-07-04
2020-07-03
2020-07-02
2020-07-01
2020-06-30
I've run the code with various date/times manually, and I cannot recreate the bug... and yet it happens once every few days in production.
This is just vanilla PHP, but it is being run as part of a Laravel project, should that factor into things. (The app timezone is set to "Europe/London".)
I'm not keen on how you're defining $start and $end. If I'm not mistaken, if the server clock happens to tick to the next second between the two variables being defined, then your interval will be 3 days, 23 hours, 59 minutes, 59 seconds - instead of exactly 4 days. This messes up your definition of $diff->days to be 3 instead of 4, leading to a missing date.
I would suggest a different approach here. Specifically, start with the current date, and subtract a day the number of times you want - since this appears to be hard-coded to 5.
$date = new DateTime();
$interval = new DateInterval("P1D");
for( $i=0; $i<5; $i++) {
$date->sub($interval);
echo $date->format("Y-m-d")."\n";
}
That $i<5 can, of course, be refactored to $i < DAYS for some appropriate constant definition, to avoid the "magic number" and allow for changing in future development.
With DateTime (and also strtotime) it is possible to process expressions like 'Today -3 Days'. Today is always 00:00 today. The calculation can be simplified as follows:
$days = 5;
for( $i=1; $i<=$days; $i++) {
echo date_create("today -$i days")->format('Y-m-d')."<br>\n";
}
I'm really not grasping how dates and times get formatted in PHP for use in mathematical equations. My goal is this;
Get a date and time from the database;
// Get array for times in
$sth = mysqli_query($conn,"SELECT * FROM ledger ORDER BY ID");
$timeins = array();
while($r = mysqli_fetch_assoc($sth)) {
$timeins[] = $r["timein"];
//OR
array_push($timeins, $r['timein']);
}
And then find the distance between the current time and the variable in the array, $timeins[0], and then put the minutes, hours, and days difference in separate simple variables for later use. These variables will be used on their own in if statements to find out if the person has passed certain amounts of time.
edit: the format of the dates being returned from the DB is in the default TIMESTAMP format for MySQL. E.g. 2018-08-06 17:38:37.
It is also possible to perform datetime operations in SQL, to get a difference between two datetime/timestamp values in days, hours, minutes... We can use expressions in the SELECT list, to return the results as columns in the resultset.
Ditching the SELECT * pattern, and specifying an explicit list of expressions that we need returned:
$sql = "
SELECT t.id
, t.timein
, TIMESTAMPDIFF(DAY ,t.timein,NOW()) AS diff_days
, TIMESTAMPDIFF(HOUR ,t.timein,NOW()) AS diff_hours
, TIMESTAMPDIFF(MINUTE ,t.timein,NOW()) AS diff_minute
FROM ledger t
ORDER BY t.id ";
if( $sth = mysqli_query($conn,$sql) ) {
// execution successful
...
} else {
// handle sql error
}
You should use the DateTime class in PHP to do any date manipulation. You can convert a string representation of a MySQL format time to a PHP DateTime object using
$date = DateTime::createFromFormat('Y-m-d H:i:s', $mysqldate);
Also you can create a DateTime object representing the current time using the constructor with no argument:
$now = new DateTime();
To get the difference between two dates as a DateInterval object, use the builtin diff method:
$diff = $now->diff($date);
As a complete example:
$now = new DateTime();
$mysqldate = '2018-04-03 12:30:01';
$date = DateTime::createFromFormat('Y-m-d H:i:s', $mysqldate);
$diff = $now->diff($date);
$diff_days = (int)$diff->format('%a');
$diff_hours = $diff->h;
$diff_minutes = $diff->m;
echo "$diff_days days, $diff_hours hours, $diff_minutes minutes";
Output:
125 days, 9 hours, 4 minutes
Note that you have to use $diff->format('%a') rather than $diff->d to get the days between two dates, as $diff->d will not include the days in any months between the two dates (in this example it will return 3 for today being August 6).
Using the DateTime Class in php is the best way to get accurate results.
$dateNow = new DateTime('now');
$dateIn = DateTime::createFromFormat('d-m-Y H:i:s', $timeins[0]);
$interval = $dateNow->diff($dateIn);
echo $interval->format('%d days, %h hours, %i minutes, %s seconds');
$deltaDays = $interval->d;
$deltaHours = $interval->h;
...
You have to make sure the input format for you DB date is correct, in this case, I assumed d-m-y H:i:s, and then you can output in any format you need as well, as shown in the date docs: http://php.net/manual/en/function.date.php
I have a database with different workdates, and I have to make a calculation that generates more dates based on a weekinterval (stored in the database) and the (in the database stored) days on which the workdays occur.
What my code does now is the following:
Read the first two workdates -> Calculate the weeks inbetween and save the week interval
Read all the workdates -> fill in the days on which a workdate occurs and save it in a contract.
Generate workdates for the next year, based on the week interval.
The point is: for each week with a week interval of 1, more days of the week should be saved as a workdate. I've used this code to do this, but it doesn't work.
// Get the last workdate's actdate.
$workdate_date = $linked_workdate['Workdate']['workdate_actdate'];
// Calculate the new workdate's date
$date = date("Y-m-d", strtotime($workdate_date . "+" . $interval . " week"));
// If 'Monday' is filled in for this contract, calculate on which day the
// Monday after the last interval is. Same for each day, obviously.
// The days are boolean.
if ($contract['Contract']['contract_maandag'] = 1){
$date = date("Y-m-d", strtotime($date, "next Monday"));
}
if ($contract['Contract']['contract_dinsdag'] = 1){
$date = date("Y-m-d", strtotime($date, "next Tuesday"));
}
// After this, save $date in the database, but that works.
Here is the error that i get:
strtotime() expects parameter 2 to be long, string given
I'm quite stuck right now, so help is appreciated!
if ($contract['Contract']['contract_maandag'] = 1){
if ($contract['Contract']['contract_dinsdag'] = 1){
This won't work. You're doing an assignment (=), so it's always true. But you want a comparison (===). It is recommended to do always (except required otherwise) to use strict (===) comparison.
Well, the = doesn't seem to be the problem, since the error is about the part that's after the comparison. Try
strtotime("$date next Monday");
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
I am aware this topic is pretty exhausted, but obviously not quite enough!
$temp_d1 = new DateTime(date('Y-m-d', $fromTime)); // 2012-01-01
$temp_d2 = new DateTime(date('Y-m-d', $endTime)); // 2013-02-01
$interval = $temp_d2->diff($temp_d1);
$monthsAhead = $interval->format('%m'); // returns 1, but I am expecting 13
How do you calculate the number of months between two dates without wrapping within a 12 month scale?
I was confusing exactly what:
$monthsAhead = $interval->format('%m');
does.
Obviously, format('%m') is just formatting the month component of the DateInterval object, not necessarily 'give me the interval as a number of months'.
In my case, I was looking for/to do this:
$monthsAhead = $interval->m + ($interval->y * 12);
http://www.php.net/manual/en/class.dateinterval.php
Hope this helps other fools in the future!