Datetime php minus one day

Php php dateinterval minus one day

How to set limit on input type date on today and day before [duplicate]

Am trying to limit the date to today’s date and a day before

Select Date: " min=? name="date"> 

You can use date time objects and modify . This is my personally preferred way (though there are others).

modify("-1 days"); $max = new DateTime(); ?> Select Date: " min=format("Y-m-d")?> max=format("Y-m-d")?> name="date"> 

You can limit the dates using the min and max attributes on the input. If you want to do this automatically with a bit of php you could try something similar to the following:

sub(new DateInterval('P1D')); // go back one day echo $yesterday->format('Y-m-d') . "\n"; ?> Select Date: " min="" max="" name="date"> 

Date Difference in php on days?, Teams. Q&A for work. Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams

PHP: determination of the date minus weekends and holidays

There is a WordPress and ACF field with a date in the format 2022-04-30. From this date need to calculate 2 other dates, -2 days and +25 days. The problem is that it is necessary to take into account only working days, i.e. check weekends and holidays.

For example, we set the date May 3, 2022, which is Tuesday. From this date, I need to subtract 2 days, i.e. May 1, 2022, but this is Sunday, so we have to return to the first working day before May 1, i.e. Friday April 29, 2022. It’s the same with holidays.

At the moment I have this code:

$setDate = get_field('set_date'); // ACF field, set May 3, 2022 (2022-05-03) $offDate = wp_date('j F Y', strtotime('-2 days', strtotime($setDate))); echo $offDate; // returns Sunday, May 1, 2022 

I found holidays and weekends in json https://github.com/d10xa/holidays-calendar/blob/master/json/consultant2022.json

So I need to compare the given date with the dates from json and if there is a match, then minus one day and check the received date again. If there is a match, again minus one day and so on until no matches are found. Am I thinking correctly and can you tell me how to implement it? I am a very bad programmer, but there is a task)

At the moment I was only able to compare the dates and return the result found/not found. But I can’t figure out how to take days off on the basis of this and send it for verification again(

$setDate = '2022-05-01'; $file = file_get_contents('https://raw.githubusercontent.com/d10xa/holidays-calendar/master/json/consultant2022.json', true); $data = json_decode($file); $found = array_search($setDate, $data->holidays); if ($found === False) < echo 'Not Found'; >else

The following has been tested on a few dates and I think it works as it should.

/* JSON file is saved locally to a sub-directory for the current working script directory. This is to avoid unneccessary hits to the remote site. */ $format='Y-m-d'; $url='https://raw.githubusercontent.com/d10xa/holidays-calendar/master/json/consultant2022.json'; $setDate = '2022-05-01'; $filepath=sprintf('%s/json/%s', __DIR__, basename( $url ) ); if( !file_exists( $filepath ) ) < file_put_contents( $filepath, file_get_contents( $url ) ); ># read the file and generate JSON $json=json_decode( file_get_contents( $filepath ) ); $hols=$json->holidays; # create the initial DateTime object and find which weekday we are dealing with # where 1 (for Monday) through 7 (for Sunday) $obj=new DateTime( $setDate ); $date=$obj->format( $format ); $day=$obj->format('N'); # Is the given date a holiday/weekend? if( array_search( $date, $hols ) ) < if( $day >5 )< # Date is either Sat or Sun. go back to previous working day $subtract = 2 - ( 7 - $day ); $int=new DateInterval( sprintf('P%sD', $subtract ) ); $obj=new DateTime( $setDate ); $previous=$obj->sub( $int ); >else< $previous=$obj->sub( new DateInterval('P2D') ); > # create the future date ( add 25 days ) $int=new DateInterval('P25D'); $obj=new DateTime( $setDate ); $future=$obj->add( $int ); if( array_search( $future->format( $format ), $hols ) )< # Future date is a holiday. go back to previous working day $day=$future->format('N'); $subtract = 2 - ( 7 - $day ); $int=new DateInterval( sprintf('P%sD',$subtract ) ); $future=$future->sub( $int ); > >else< # Given date is NOT a holiday. # take a copy of the original DateTime object for generating future date. $ref=new DateTime( $setDate ); $int=new DateInterval( 'P2D' ); $previous=$obj->sub( $int ); $day=$previous->format('N'); # Is this a holiday? if( $day > 5 )< # yes - go back to previous working day $subtract = 2 - ( 7 - $day ); $int=new DateInterval( sprintf('P%sD',$subtract ) ); $previous=$previous->sub( $int ); > $int=new DateInterval('P25D'); $future=$ref->add( $int ); $day=$future->format('N'); # Is this a holiday? if( $day > 5 )< $subtract = 2 - ( 7 - $day ); $int=new DateInterval( sprintf('P%sD',$subtract ) ); $future=$future->sub( $int ); > > printf( ' 
Given date: %s Previous (-2): %s Future (+25): %s
 
 ', $date, $previous->format( $format ), $future->format( $format ) ); 
 
Given date: 2022-05-01 Previous (-2): 2022-04-29 Future (+25): 2022-05-26 
 
Datetime — How to Minus two dates in php, 60*60*24 is second of days. diff result is 604800 about betweeen two date seconds. You have to divide to day. First 60 is min, second 60 hour and last 24 is for the day. So 60*60*24 = 86400 is mean one (1) date. – Bora. Aug 23, 2013 at 6:47. How many days difference is from 2013 …
 
PHP minus one week Sunday
 
I’m a newbie in both php and codeigniter.
 
I have a function which is used to label the x-axis of a chart of weekly sums of kilometers rode on a bicycle. It works (or is supposed to work) by counting back from today’s date either zero Sundays or 1, 2, 3, etc as passed by an array.
 
I am using PHP Version 5.6.19, and this is my code:
 
 public function week_start($date, $counter)
 
$datenumbers = array(0, 1, 2, 3, 4, 5, 6, 7); $dateResult = $this->bike_model->week_start(date("Y/m/d"), $datenumber); 
 
 for ($i = 0; $i < 8; $i++)< echo ""; echo $dates_for_table[$i];> 
 
I have had it up for few days and it has been showing Sunday July 3 as the first number on on my x-axis, as it should have. But today, it is Sunday and I expect it to now say Sunday July 10 as my most recent week. However, it is still showing last sunday!
 
So my question is, is there a bug here or am I misunderstanding how ‘- one week Sunday’ should work?
 
My goal is to have my function with zero as the argument return the most recent week start—either the current Sunday if it is Sunday or the past sunday. I have found a workaround by adding an if else to the function checking if it is Sunday.
 
if ($tstring == strtotime('this Sunday')) < $wknumber = $counter;>else
 
However, this feels clunky and I’m looking for a more elegant way. I also want to understand what went wrong with «- one week Sunday» since I thought I had understood how it was working.
 
Here is how I did something similar using the object oriented DateTime style:
 
$startDate = new DateTime('NOW'); // can pass date string if not today # reset start date to last Sunday or today if it is Sunday: $startDate->sub(new DateInterval('P'.($startDate->format('w')).'D')); # for debugging, output first Sunday: echo $startDate->format('Y-m-d H:i:s') . " => full date and time (check for timezone issues) \n\n"; $weeks = 8; for ($i = 0; $i < $weeks; $i++) < # don't decrement first pass; sub 1 week on subsequent: $startDate->sub(new DateInterval('P'. (($i) ? 1 : 0) .'W')); echo $startDate->format('Y-m-d') . "\n"; >; 
 
You should be able to adapt for your purpose.
 
If you want to get fancy, you can also use DatePeriod instead of using sub within a loop:
 
$period = new DatePeriod( $startDate, DateInterval::createFromDateString('-1 week'), 7 ); foreach ($period as $date) < echo $date->format('Y-m-d') . "\n"; > 
 
Calculate total seconds in PHP DateInterval, It always feels nice when what you need is already present as a class. In your case, you need a total number of seconds. Why not make a class out of that? The code could look like that: (new TotalFullSeconds( new IntervalFromRange( new DateTimeFromISO8601String(‘2017-07-03T14:27:39+00:00’), new …
 
How to minus one day from date using date() in phph
 
php date format minus 1 day
 
date('Y-m-d',(strtotime ( '-1 day' , strtotime ( $date) ) ));
 
Php — Adding one day to a date, My code to add one day to a date returns a date before day adding: 2009-09-30 20:24:00 date after adding one day SHOULD be rolled over to the next month: 1970-01-01 17:33:29 
 
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PHP: Subtract days from a date.
 
In this tutorial, we will show you how to subtract days, weeks, or months from a date using PHP.
 
To do this, we can use the strtotime function or the DateTime object.
 
How to subtract one day from a date using PHP.
 
To get yesterday’s date, you can pass in the string “yesterday” to the strtotime function like so:
 
//Yesterday $yesterday = date("Y-m-d", strtotime("yesterday")); //On 2019-08-19, this resulted in 2019-08-18 echo $yesterday;
 
Passing in the string “yesterday” to strtotime is basically the exact same thing as subtracting “-1 day” from today’s date.
 
Here is a DateTime alternative, which you can use if your PHP version is 5.3.0 or above:
 
//New DateTime object representing today's date. $currentDate = new DateTime(); //Use the sub function to subtract a DateInterval $yesterdayDT = $currentDate->sub(new DateInterval('P1D')); //Get yesterday's date in a YYYY-MM-DD format. $yesterday = $yesterdayDT->format('Y-m-d'); //Print it out. echo $yesterday;
 
PHP introduced the DateInterval object in version 5.3.0. This object represents a date interval.
 
In the example above, we set the DateInterval object to P1D, which means a period of one day. We then subtracted that interval from the current date.
 
If you would like to subtract five days instead of one day, then you can use P5D instead of P1D.
 
If you want to subtract a day from a given date:
 
//Pass the date you want to subtract from in as //the $time parameter for DateTime. $currentDate = new DateTime('2019-01-01'); //Subtract a day using DateInterval $yesterdayDT = $currentDate->sub(new DateInterval('P1D')); //Get the date in a YYYY-MM-DD format. $yesterday = $yesterdayDT->format('Y-m-d'); //Print it out. echo $yesterday;
 
In the example above, we subtracted one day from 2019-01-01. This resulted in “2018-12-31”.
 
Subtract 7 days.
 
To get last week’s date in PHP, we can use the strtotime function like so:
 
//7 days ago - last week. $lastWeek = date("Y-m-d", strtotime("-7 days")); //On 2019-08-19, this resulted in 2019-08-12 echo $lastWeek;
 
In this case, we passed “-7 days” in as a parameter to the strtotime function.
 
This works because the strtotime function is pretty good at parsing textual date descriptions.
 
Here is an alternative solution for those of you who prefer to use the DateTime object:
 
//Today's date. $currentDate = new DateTime(); //Subtract a day using DateInterval $lastWeekDT = $currentDate->sub(new DateInterval('P1W')); //Get the date in a YYYY-MM-DD format. $lastWeek = $lastWeekDT->format('Y-m-d'); //Print out the result. echo $lastWeek;
 
Above, we set “P1W” as the $interval_spec parameter for DateInterval. This translates into “a period of one week.”
 
If you want to subtract two weeks instead of one, then you can change “P1W” to “P2W”.
 
Last month / -30 days.
 
Getting last month’s date can be tricky. This is because the number of days in a month can vary.
 
For example, the month of October has 31 days, while September has 30.
 
Take the following example:
 
//Last month $lastMonth = date("Y-m-d", strtotime("-1 month")); //On 2019-08-19, this resulted in 2019-07-19 echo $lastMonth;
 
The snippet above will work in most situations.
 
However, on October 31st, the function will actually return “October 1st” instead of “September 30th”.
 
This is because PHP’s strtotime function treats “-1 month” the same as “-30 days.”
 
If you want to get the previous month’s name or number, then you will need to use the following code instead:
 
//Get the first date of the previous month. echo date("Y-m-d", strtotime("first day of previous month"));
 
If you run the script above on October 31st, it will return September 1st.
 
As you can see, dealing with dates can be very tricky.
 
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