Related
I can add any number of months to a date:
strtotime("+ 1 months", '2017-01-31') // result: 2017-03-03
But I want to do this without going to the next month.
in this case I want the result 2017-02-28, that is, the last day of the month before the target month.
There seems to be a lot of overcomplicating in these answers. You want the last day of the month before your target month, which is also always going to be 1 day before the first day of the target month. This can be expressed quite simply:
$months = 1;
$relativeto = '2017-01-31';
echo date(
'Y-m-d',
strtotime(
'-1 day',
strtotime(
date(
'Y-m-01',
strtotime("+ $months months", strtotime($relativeto))
)
)
)
);
Try using the DateTime object like so:
$dateTime = new \DateTime('2017-01-31');
$dateTime->add(new \DateInterval('P1M'));
$result = $dateTime->format('Y-m-d');
Use PHP's DateTime to accomplish this, specifically the modify() method.
$date = new DateTime('2006-12-12');
$date->modify('+1 month');
echo $date->format('Y-m-d');
If the idea here is not to allow the date to overflow into the next month, which PHP does, then you'll have to impose that constraint in your logic.
One approach is to check the modified month against the given month before returning the updated date.
function nextMonth(DateTimeImmutable $date) {
$nextMonth = $date->add(new DateInterval("P1M"));
$diff = $nextMonth->diff($date);
if ($diff->d > 0) {
$nextMonth = $nextMonth->sub(new DateInterval("P{$diff->d}D"));
}
return $nextMonth;
}
$date = new DateTimeImmutable("2017-01-31");
$nextMonth = nextMonth($date);
echo "{$date->format('Y-m-d')} - {$nextMonth->format('Y-m-d')}\n";
//2017-01-31 - 2017-02-28
$date = new DateTimeImmutable("2017-10-31");
$nextMonth = nextMonth($date);
echo "{$date->format('Y-m-d')} - {$nextMonth->format('Y-m-d')}\n";
//2017-10-31 - 2017-11-30
The nextMonth() function in the example above, imposes the constraint of not overflowing into the next month. Note that what PHP actually does is try to find the corresponding day, in the following consecutive number of months added, not just add a given number of months to the existing date. I simply undo this last part by subtracting any additional days beyond the 1 month interval in the function above.
So for example, strtotime("+1 month") for the date "2017-01-31", tells PHP find the 31st day that is +1 month from "2017-01-31". But of course, there are only 28 days in February, so PHP goes into March and counts up 3 more days to compensate.
In other words, it's not just add +1 month to this date, but add +1 month to this date and arrive at the same day of the month as is in the given date. Which is where the overflow happens.
Update
Since you've now made it clear that you actually want the last day of the month (not necessarily the same day of the next month without the overflow provision) you should instead just explicitly set the day of the month.
function nextMonth(DateTimeImmutable $date) {
$nextMonth = $date->setDate($date->format('Y'), $date->format('n') + 1, 1);
$nextMonth = $date->setDate($nextMonth->format('Y'), $nextMonth->format('n'), $nextMonth->format('t'));
return $nextMonth;
}
$date = new DateTimeImmutable("2017-02-28");
$nextMonth = nextMonth($date);
echo "{$date->format('Y-m-d')} - {$nextMonth->format('Y-m-d')}\n";
// 2017-02-28 - 2017-03-31
suppose I have an initial date whose year was prior to that of the current year and I want to repeat the event every 7 days but only in the current year.
How would I find the first occurrence in the current year?
I realize I can do it with a loop like this:
$reOccurringEvent =new DateTime('2013-12-01');
$interval = new DateInterval('P7D');
while($reOccurringEvent->format('Y') < date('Y') ){
$reOccurringEvent->add($interval);
}
echo $reOccurringEvent->format('d m Y'); //05 01 2014
But it strikes me there should be a more efficient way to achieve this rather than repeatedly adding an interval to the date (it would happen many times if the initial date was some years ago).
I was hoping to be able to calculate the number of times the interval should be added and just do it a single time.
I was thinking something like:
$date = new DateTime();
$diff = $date->diff($reOccurringEvent)->days%7;
But obviously that doesn't work and I can't quite figure out the logic of how to do it.
More generically, the algorithm would be to find the number of intervals between the given date and the last day of last year. Then multiplying the interval by the number of intervals + 1 to get the first interval of the current year.
$date1="12/9/2013";
$ts1 = strtotime($date1);
$ts2 = strtotime("12/31/" . Date("Y")-1);
//get the number of seconds between the date and first of the year
$seconds_diff = $ts2 - $ts1;
echo "$seconds_diff <br>";
//get the number of days
$dayDiff=$seconds_diff/86400;
//how many intervals?
$intervalDays = "10";
//get the number of intervals from start date to last day of last year
$numIntervals = floor($dayDiff/$intervalDays);
echo $numIntervals."<br>";
//now the total intervals to get into the current year is one more interval, turn this into days
$totIntervals= ($numIntervals* $intervalDays)+$intervalDays;
//Date Time date in question
$theDt = new DateTime($date1);
//Add the intervals we calculated to the date in question, and we have the first date of the interval for the current year...
$theDt->add(new DateInterval('P' . $totIntervals. 'D'));
echo "The first date of the intreval is: " . $theDt->format('Y-m-d');
I think, if you are doing 7 day intervals, you can find out the Day of week of your initial date, and then get the first date of the current year with that day of week...
Find out day of week: How to find the day of week from a date using PHP?
Find out date with that day of week for this year: Getting first weekday in a month with strtotime
Putting it together:
$date=Date("2/8/2012");
//Get the day of week for the date in question
$dayOfWeek = date('l', strtotime($date));
echo "The day of week for the given date is: $dayOfWeek <br>";
//Get the current year
$thisYear = date("Y");
echo "This year: $thisYear <br>";
//Create a date with the first occurence of the day of week of the given date for the current year
$firstOccurenceThisYear = date("m/d/y", strtotime("January " .$thisYear ." " . $dayOfWeek));
echo "The first interval of the year is: $firstOccurenceThisYear";
/*
Output:
This year: 2014
The day of week for the given date is: Wednesday
The first interval of the year is: 01/01/14
*/
Here is a slightly modified version of #Dan's second answer which worked well for me.
Benchmarks shown below.
$date="1985-02-18";
$intervalDays = "5";
//original version
$benchMark = microtime(true);
$dt1 = new DateTime($date);
$interval = new DateInterval("P{$intervalDays}D");
while ($dt1->format('Y') < date('Y')) {
$dt1->add($interval);
}
echo $dt1->format('d m Y') . '<br>';
echo microtime(true)-$benchMark.'<br>';
//new version
$benchMark = microtime(true);
$dt1 = new DateTime($date);
$dt2 = new DateTime("12/31/" . ((int) Date("Y") - 1));
$dayDiff = $dt1->diff($dt2)->days;
$numIntervals = floor($dayDiff / $intervalDays);
$totIntervals = ($numIntervals * $intervalDays) + $intervalDays;
$dt1->add(new DateInterval('P' . $totIntervals . 'D'));
echo $dt1->format('d m Y').'<br>';
echo microtime(true)-$benchMark.'<br>';
exit;
output
02 01 2014
0.0145111083984
02 01 2014
0.000123977661133
I have a dynamic date, now what i want is that finding the date after exact one week, i have achieved that with the code below, but now i want that now many days are left for that week after date to come. i have got some sort of time stamp, but i don't know how to convert it to DAYS LEFT.
$weekDate = date( "d/m/Y", strtotime("19-05-2014") + 86400 * 7 );
echo $weekDate;// THATS PERFECT
////////////////////////////////////////////////////////////////
$future = strtotime( $weekDate ); //Future date.
$datediff = time() - $future;
$days = floor( ( ( $datediff / 24 ) / 60 ) / 60 ); //this is not perfect, returns some
sort of timestamp
I have tried other methods which are fine, but if week completes on 26, and today is 25th it gives me 0 days left, but it should say 1 day left. please help me.
In your $date_diff now is less than the future date thats why its zero. Inside strtotime() function, you can directly put a relative date inside. In this case, for one week you can use +1 week or +7 days. Consider this example:
$next_week = date('d/m/Y', strtotime('19-05-2014 +1 week')); // 26/05/2014
$next_week = strtotime('19-05-2014 +7 days');
$difference = $next_week - time(); // next weeks date minus todays date
$difference = date('j', $difference);
echo $difference . (($difference > 1) ? ' days ' : ' day ') . ' left';
// should output: 1 day left
Alright. I did something. Here's the code
$startDate = strtotime("19-05-2014");
$endDate = $startDate + 604800;
$diff = ($endDate - time()) / 60 / 60 / 24;
if ($diff < 1 && $diff > 0) {
$days = 1;
} else {
$days = floor($diff);
}
echo $days;
The problem you have with getting "1 day" if the date is tomorrow is the floor method. strtotime() gives you the time at 0 a.m. if you don't set it by your own. Because of that the difference between now and tomorrow is less than 1 which is 0 if you floor that. I created an if-clause for that.
But that will give you "1 day" for today and "1 day" for yesterday (last 2 days before the final date). If you want that better, you have to specify time in your initial date (19-05-2014).
Use DateTime for date and time calculations.
$weekDate = new \DateTime('+ 1 week');
$future = new \DateTime('+ 3 days');
$daysLeft = $weekDate->diff($future)->days;
echo $daysLeft; //4
See it working.
Reference http://php.net/datetime
I'm trying to accomplish the following task. I'm developing a custom calendar with three views (day,week and month), there may be something out there already but I'm rewriting this as part of learning tool for me as well.
So user will face with Day view when they first visit, with arrows to go back and forth to next day or previous day of course. If they click on Week View, it will give them a 7 days overview with today date as default, and once again they can go back and forth to next week or previous week. The last view is the full month calendar, once they click on the day, it will give them the detail of the day and at the same time reset the default as the day they pick. So if they go back to week view, they will see the detail for the week contain the day that they picked. This is where I have trouble wrapping my head around with, I know there are PHP functions that determine the day of the week but I can't seem to think about how to pass in the date and get the full week starting from Sunday for the day that passed in. For example, if I passed in 10/12/2012, I'd like to start the week at 10/07/12 - 10/13/12.
Thank you kindly for your help or pointing to the right direction. Please excuse my grammar/spelling mistakes as well.
Assuming that $selection represents the user's selected date as an
integer timestamp, date('N', $selection) will return a numeric
representation of day of the week of their selection (e.g. Monday =
1 through Sunday = 7).
This result also represents the number of days since the start of
the selected week (the previous Sunday) - $offset. Of course, if
you're beginning with a string representation of the users selected
date (as in your question, 10/12/2012), you would first need to
convert the date to an integer timestamp.
$selection = strtotime($selection); //if $selection is in string format
$offset = date('N', $selection);
Now you may use $offset to establish date of the the start of the
week (the previous Sunday) - $weekstart.
$weekstart = strtotime("$selection -$offset day");
Once you've got the start of the week, the end of the week
($weekend) is, of course, 6 days later, but to calculate this
date, you first need to convert $weekstart from an integer
timestamp into a string representation of the date. You'll also need to convert the result ($weekend) into a string representation of the date.
$weekstart = stringftime("%m/%d/%Y", $weekstart); //date format = 10/07/2012
$weekend = strtotime("$weekstart +6 day");
$weekend = stringftime("%m/%d/%Y", $weekend);
So the following:
$selection = "10/12/2012";
$selection = strtotime($selection);
$offset = date('N', $selection);
$weekstart = strtotime("$selection -$offset day");
$weekstart = stringftime("%m/%d/%Y", $weekstart);
$weekend = strtotime("$weekstart +6 day");
$weekend = stringftime("%m/%d/%Y", $weekend);
$output = "Selected Date = $selection \n Selected Week = $weekstart - $weekend";
echo $output;
results in:
Selected Date = 10/12/2012
Selected Week = 10/07/2012 - 10/13/2020
See:
http://php.net/manual/en/function.date.php
http://php.net/manual/en/function.strtotime.php
http://php.net/manual/en/function.strftime.php
use strtotime() function coupled with date() function for this task
for eg. 1 day before 2012-10-09 is
echo date('Y-m-d', strtotime("2012-10-09 -1 day"));
Building on Wisdom's answer, you can find the 1st day of the week in a couple steps. Check out the PHP date() manual for more options, but I believe the following code will work:
// figure out how many days back you have to go, to get to Sunday
$d = date('N', strtotime($mydate));
// figure out Sunday's date
$beginning_of_week = date('Y-m-d', strtotime($mydate." -{$d} days"));
$end_of_week = date('Y-m-d', strtotime($beginning_of_week." +1 week"));
echo "The week {$beginning_of_week} to {$end_of_week}! ";
Try this function (source: Marty Wallace).
The date you are interested in is $date (in YYYY-MM-DD format), and $rollover in full day format (e.g. Friday).
function getWeeks($date, $rollover)
{
$cut = substr($date, 0, 8);
$daylen = 86400;
$timestamp = strtotime($date);
$first = strtotime($cut . "00");
$elapsed = ($timestamp - $first) / $daylen;
$i = 1;
$weeks = 1;
for($i; $i<=$elapsed; $i++)
{
$dayfind = $cut . (strlen($i) < 2 ? '0' . $i : $i);
$daytimestamp = strtotime($dayfind);
$day = strtolower(date("l", $daytimestamp));
if($day == strtolower($rollover)) $weeks ++;
}
return $weeks;
}
I've seen some variants on this question but I believe this one hasn't been answered yet.
I need to get the starting date and ending date of a week, chosen by year and week number (not a date)
example:
input:
getStartAndEndDate($week, $year);
output:
$return[0] = $firstDay;
$return[1] = $lastDay;
The return value will be something like an array in which the first entry is the week starting date and the second being the ending date.
OPTIONAL: while we are at it, the date format needs to be Y-n-j (normal date format, no leading zeros.
I've tried editing existing functions that almost did what I wanted but I had no luck so far.
Using DateTime class:
function getStartAndEndDate($week, $year) {
$dto = new DateTime();
$dto->setISODate($year, $week);
$ret['week_start'] = $dto->format('Y-m-d');
$dto->modify('+6 days');
$ret['week_end'] = $dto->format('Y-m-d');
return $ret;
}
$week_array = getStartAndEndDate(52,2013);
print_r($week_array);
Returns:
Array
(
[week_start] => 2013-12-23
[week_end] => 2013-12-29
)
Explained:
Create a new DateTime object which defaults to now()
Call setISODate to change object to first day of $week of $year instead of now()
Format date as 'Y-m-d' and put in $ret['week_start']
Modify the object by adding 6 days, which will be the end of $week
Format date as 'Y-m-d' and put in $ret['week_end']
A shorter version (works in >= php5.3):
function getStartAndEndDate($week, $year) {
$dto = new DateTime();
$ret['week_start'] = $dto->setISODate($year, $week)->format('Y-m-d');
$ret['week_end'] = $dto->modify('+6 days')->format('Y-m-d');
return $ret;
}
Could be shortened with class member access on instantiation in >= php5.4.
Many years ago, I found this function:
function getStartAndEndDate($week, $year) {
$dto = new DateTime();
$dto->setISODate($year, $week);
$ret['week_start'] = $dto->format('Y-m-d');
$dto->modify('+6 days');
$ret['week_end'] = $dto->format('Y-m-d');
return $ret;
}
$week_array = getStartAndEndDate(52,2013);
print_r($week_array);
We can achieve this easily without the need for extra computations apart from those inherent to the DateTime class.
function getStartAndEndDate($year, $week)
{
return [
(new DateTime())->setISODate($year, $week)->format('Y-m-d'), //start date
(new DateTime())->setISODate($year, $week, 7)->format('Y-m-d') //end date
];
}
The setISODate() function takes three arguments: $year, $week, and $day respectively, where $day defaults to 1 - the first day of the week. We therefore pass 7 to get the exact date of the 7th day of the $week.
Slightly neater solution, using the "[year]W[week][day]" strtotime format:
function getStartAndEndDate($week, $year) {
// Adding leading zeros for weeks 1 - 9.
$date_string = $year . 'W' . sprintf('%02d', $week);
$return[0] = date('Y-n-j', strtotime($date_string));
$return[1] = date('Y-n-j', strtotime($date_string . '7'));
return $return;
}
shortest way to do it:
function week_date($week, $year){
$date = new DateTime();
return "first day of the week is ".$date->setISODate($year, $week, "1")->format('Y-m-d')
."and last day of the week is ".$date->setISODate($year, $week, "7")->format('Y-m-d');
}
echo week_date(12,2014);
You can get the specific day of week from date as bellow that I get the first and last day
$date = date_create();
// get the first day of the week
date_isodate_set($date, 2019, 1);
//convert date format and show
echo date_format($date, 'Y-m-d') . "\n";
// get the last date of the week
date_isodate_set($date, 2019, 1, 7);
//convert date format and show
echo date_format($date, 'Y-m-d') . "\n";
Output =>
2018-12-31
2019-01-06
The calculation of Roham Rafii is wrong. Here is a short solution:
// week number to timestamp (first day of week number)
function wn2ts($week, $year) {
return strtotime(sprintf('%dW%02d', $year, $week));
}
if you want the last day of the week number, you can add up 6 * 24 * 3600
This is an old question, but many of the answers posted above appear to be incorrect.
I came up with my own solution:
function getStartAndEndDate($week, $year){
$dates[0] = date("Y-m-d", strtotime($year.'W'.str_pad($week, 2, 0, STR_PAD_LEFT)));
$dates[1] = date("Y-m-d", strtotime($year.'W'.str_pad($week, 2, 0, STR_PAD_LEFT).' +6 days'));
return $dates;
}
First we need a day from that week so by knowing the week number and knowing that a week has seven days we are going to do so the
$pickADay = ($weekNo-1) * 7 + 3;
this way pickAday will be a day in our desired week.
Now because we know the year we can check which day is that.
things are simple if we only need dates newer than unix timestamp
We will get the unix timestamp for the first day of the year and add to that 24*3600*$pickADay and all is simple from here because we have it's timestamp we can know what day of the week it is and calculate the head and tail of that week accordingly.
If we want to find out the same thing of let's say 12th week of 1848 we must use another approach as we can not get the timestamp. Knowing that each year a day advances 1 weekday meaning (1st of november last year was on a sunday, this year is on a monday, exception for the leap years when it advances 2 days I believe, you can check that ). What I would do if the year is older than 1970 than make a difference between it and the needed year to know how many years are there, calculate the day of the week as my pickADay was part of 1970, shift it back one weekday for each. $shiftTimes = ($yearDifference + $numberOfLeapYears)%7, in the difference. shift the day backwords $shiftTimes, then you will know what day of the week was that day those years ago, then find the weekhead and weektail. Same thing can be used also for the future if it seems simpler. Try it if it works and tell me if it does not.
For documentation (since Google ranks this question first when searching for "php datetime start end this week").
If you need the startdate and enddate for the current week (using DateTime):
$dateTime = new DateTime('now');
$monday = clone $dateTime->modify(('Sunday' == $dateTime->format('l')) ? 'Monday last week' : 'Monday this week');
$sunday = clone $dateTime->modify('Sunday this week');
var_dump($monday->format('Y-m-d')); // e.g. 2018-06-25
var_dump($sunday->format('Y-m-d')); // e.g. 2018-07-01
Hope this will help.
The "first day of the week" is subjective. Some cultures use "Monday" others "Sunday", maybe others something else?
For my purposes, I want the first day of the week to be "Sunday" and the last day of the week to be "Saturday".
Also, using DateTime with no arguments will default to "now" which includes the current time. The following method will disregard the current time by specifying "today" in the DateTime constructor.
Furthermore the string "sunday this week" does not seem to be reliable. It actually will return Sunday the next week (according to my view of what a week is).
I've built a method which returns a PHP object containing two DateTime objects. One for the first day (Sunday) of the given week, the second for the last day (Saturday) of the given week.
function get_first_and_last_day_of_week( $year_number, $week_number ) {
// we need to specify 'today' otherwise datetime constructor uses 'now' which includes current time
$today = new DateTime( 'today' );
return (object) [
'first_day' => clone $today->setISODate( $year_number, $week_number, 0 ),
'last_day' => clone $today->setISODate( $year_number, $week_number, 6 )
];
}
Have you tried PHP relative dates? It might work.
Even if you dont want to use a specific date you cannot escape it. You can calculate a week based on the date ONLY.
Steps:
get the first day of the year
decide when the first week starts ( there are some rules that include first Thursday if I remember.
add some number of weeks (your first param). Zend_Date has an add() function where you can add weeks for example. This will give you the first day of the week.
offset and get the last day.
I would recommend working with a consistent dates sistem like Zend_Date or Pear Date.
function getStartAndEndDate($week, $year)
{
$week_start = new DateTime();
$week_start->setISODate($year,$week);
$return[0] = $week_start->format('d-M-Y');
$time = strtotime($return[0], time());
$time += 6*24*3600;
$return[1] = date('d-M-Y', $time);
return $return;
}
$dateParam = '2018-06-10';
$week = date('w', strtotime($dateParam));
$date = new DateTime($dateParam);
$firstWeek = $date->modify("-".$week." day")->format("Y-m-d H:i:s");
$endWeek = $date->modify("+6 day")->format("Y-m-d H:i:s");
echo $firstWeek."<br/>";
echo $endWeek;
will print
2018-06-10 00:00:00
2018-06-16 00:00:00
hopefully will help