PHP adding exact weekdays to a timestamp - php

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

Related

Next delivery date listed on website

I'm building a website for a business that makes deliveries every second Thursday. I need to display when the next delivery is going to be, and have that date change to two weeks forward when the previous delivery date is reached.
Based on what I've been able to research so far, I've cobbled together this code:
$start_date = '2016-10-27'; // next delivery date to start counting from
// create a DateTime object that represents start of sequence
$start_datetime = DateTime::createFromFormat('Y-m-d', $start_date);
// create a DateTime object representing the current date
$current_datetime = new DateTime('today');
$date_interval = new DateInterval('P2W'); // for delivery every 2 weeks
// determine end date for DatePeriod object that will later be used
// this is no further out than current date plus the interval
$end_datetime = new DateTime('tomorrow');
$end_datetime->add($date_interval);
$date_period = new DatePeriod($start_datetime, $date_interval, $end_datetime);
// iterate until the last date in the set
foreach($date_period as $dp) {
$next_delivery = $dp;
}
?>
<div class="header-next-delivery">
Next delivery: <?php echo $next_delivery->format('l, M j, Y'); ?>
</div>
This seems to work, but I can't help thinking that there must be a more elegant way to do this than having to iterate through a set of dates from the start date to the last date in the set. As time passes, the set will just get bigger and bigger.
Also, I'm having trouble figuring out the internal workings of these functions -- how would I set the exact time that the displayed delivery date bumps forward by two weeks?
Thanks for any insight!
You can get date interval in days from $start_date to current date. Divide it by 14 (two weeks), get remainder and substract it from 14. Then you can add that value of days to current date.
$start_date = date_create('2016-10-27');
$current_date = date_create();
$interval = date_diff($start_date, $current_date);
$days_diff = (int)$interval->format('%a');
$current_date->add(14 - ($days_diff % 14).' days');

How to get a full span of the start and end of a week

I was trying to find a simple solution to getting a date range, specifically the start of a week to the end of a week.
I found a few solutions here which were very complicated. I wanted a two line solution.
I thought my solution might help someone so I posted it.
$date = new \DateTime(/* any point in time and space */);
$week_start = new \DateTime(sprintf('%s today this week', $date->format('Y-m-d H:i:s')));
$week_end = new \DateTime(sprintf('%s today next week -1 second', $date->format('Y-m-d H:i:s')));
echo sprintf('%s -> %s', $week_start->format('Y-m-d H:i:s'), $week_end->format('Y-m-d H:i:s'));
A commonly overlooked feature of PHP's english to date expressions is how you can inherit words and dates from each other in what might appear to be conflicting expressions.
However you can add multiple expressions, and each one will inherit from the last.
So by adding a specific date and formatting it at the start of the expression, you can add today to get the start of that date.
After that this week to get the start of the week, inherits from the same, plus today, so the start of this week becomes the start of that week.
For the end date, I used the next week on the same principle, but applied -1 second afterwards, which gives us the first and last seconds of the week.
An alternative approach to the same result:-
$sunday = 0;
$monday = 1;
$date = new \DateTime(/* any point in time and space */);
$weekStart = (new \DateTime())->setISODate((int)$date->format('o'), (int)$date->format('W'), $monday)->setTime(0, 0, 0);
$weekEnd = (new \DateTime())->setISODate((int)$date->format('o'), (int)$date->format('W') + 1, $sunday)->setTime(23, 59, 59);
echo sprintf('%s -> %s', $weekStart->format('Y m d H:i:s'), $weekEnd->format('Y m d H:i:s'));
Personally, I prefer this type of method to passing a 'natural language' type string which, to me, has ambiguous meaning.

Generating a date based on a weekinterval, a day, and an existing date

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");

PHP Adding 15 minutes to Time value

I have a form that receives a time value:
$selectedTime = $_REQUEST['time'];
The time is in this format - 9:15:00 - which is 9:15am. I then need to add 15 minutes to this and store that in a separate variable but I'm stumped.
I'm trying to use strtotime without success, e.g.:
$endTime = strtotime("+15 minutes",strtotime($selectedTime)));
but that won't parse.
Your code doesn't work (parse) because you have an extra ) at the end that causes a Parse Error. Count, you have 2 ( and 3 ). It would work fine if you fix that, but strtotime() returns a timestamp, so to get a human readable time use date().
$selectedTime = "9:15:00";
$endTime = strtotime("+15 minutes", strtotime($selectedTime));
echo date('h:i:s', $endTime);
Get an editor that will syntax highlight and show unmatched parentheses, braces, etc.
To just do straight time without any TZ or DST and add 15 minutes (read zerkms comment):
$endTime = strtotime($selectedTime) + 900; //900 = 15 min X 60 sec
Still, the ) is the main issue here.
Though you can do this through PHP's time functions, let me introduce you to PHP's DateTime class, which along with it's related classes, really should be in any PHP developer's toolkit.
// note this will set to today's current date since you are not specifying it in your passed parameter. This probably doesn't matter if you are just going to add time to it.
$datetime = DateTime::createFromFormat('g:i:s', $selectedTime);
$datetime->modify('+15 minutes');
echo $datetime->format('g:i:s');
Note that if what you are looking to do is basically provide a 12 or 24 hours clock functionality to which you can add/subtract time and don't actually care about the date, so you want to eliminate possible problems around daylights saving times changes an such I would recommend one of the following formats:
!g:i:s 12-hour format without leading zeroes on hour
!G:i:s 12-hour format with leading zeroes
Note the ! item in format. This would set date component to first day in Linux epoch (1-1-1970)
strtotime returns the current timestamp and date is to format timestamp
$date=strtotime(date("h:i:sa"))+900;//15*60=900 seconds
$date=date("h:i:sa",$date);
This will add 15 mins to the current time
To expand on previous answers, a function to do this could work like this (changing the time and interval formats however you like them according to this for function.date, and this for DateInterval):
(I've also written an alternate form of the below function here.)
// Return adjusted time.
function addMinutesToTime( $time, $plusMinutes ) {
$time = DateTime::createFromFormat( 'g:i:s', $time );
$time->add( new DateInterval( 'PT' . ( (integer) $plusMinutes ) . 'M' ) );
$newTime = $time->format( 'g:i:s' );
return $newTime;
}
$adjustedTime = addMinutesToTime( '9:15:00', 15 );
echo '<h1>Adjusted Time: ' . $adjustedTime . '</h1>' . PHP_EOL . PHP_EOL;
get After 20min time and date
function add_time($time,$plusMinutes){
$endTime = strtotime("+{$plusMinutes} minutes", strtotime($time));
return date('h:i:s', $endTime);
}
20 min ago Date and time
date_default_timezone_set("Asia/Kolkata");
echo add_time(date("Y-m-d h:i:sa"),20);
In one line
$date = date('h:i:s',strtotime("+10 minutes"));
You can use below code also.It quite simple.
$selectedTime = "9:15:00";
echo date('h:i:s',strtotime($selectedTime . ' +15 minutes'));
Current date and time
$current_date_time = date('Y-m-d H:i:s');
15 min ago Date and time
$newTime = date("Y-m-d H:i:s",strtotime("+15 minutes", strtotime($current_date)));
Quite easy
$timestring = '09:15:00';
echo date('h:i:s', strtotime($timestring) + (15 * 60));

PHP DateTime credit card expiration

I'm trying to use DateTime to check if a credit card expiry date has expired but I'm a bit lost.
I only want to compare the mm/yy date.
Here is my code so far
$expmonth = $_POST['expMonth']; //e.g 08
$expyear = $_POST['expYear']; //e.g 15
$rawExpiry = $expmonth . $expyear;
$expiryDateTime = \DateTime::createFromFormat('my', $rawExpiry);
$expiryDate = $expiryDateTime->format('m y');
$currentDateTime = new \DateTime();
$currentDate = $currentDateTime->format('m y');
if ($expiryDate < $currentDate) {
echo 'Expired';
} else {
echo 'Valid';
}
I feel i'm almost there but the if statement is producing incorrect results. Any help would be appreciated.
It's simpler than you think. The format of the datess you are working with is not important as PHP does the comparison internally.
$expires = \DateTime::createFromFormat('my', $_POST['expMonth'].$_POST['expYear']);
$now = new \DateTime();
if ($expires < $now) {
// expired
}
You can use the DateTime class to generate a DateTime object matching the format of your given date string using the DateTime::createFromFormat() constructor.
The format ('my') would match any date string with the string pattern 'mmyy', e.g. '0620'. Or for dates with 4 digit years use the format 'mY' which will match dates with the following string pattern 'mmyyyy', e.g. '062020'. It's also sensible to specify the timezone using the DateTimeZone class.
$expiryMonth = 06;
$expiryYear = 20;
$timezone = new DateTimeZone('Europe/London');
$expiryTime = \DateTime::createFromFormat('my', $expiryMonth.$expiryYear, $timezone);
See the DateTime::createFromFormat page for more formats.
However - for credit/debit card expiry dates you will also need to take into account the full expiry DATE and TIME - not just the month and year.
DateTime::createFromFormat will by default use todays day of the month (e.g. 17) if it is not specified. This means that a credit card could appear expired when it still has several days to go. If a card expires 06/20 (i.e. June 2020) then it actually stops working at 00:00:00 on 1st July 2020. The modify method fixes this. E.g.
$expiryTime = \DateTime::createFromFormat('my', $expiryMonth.$expiryYear, $timezone)->modify('+1 month first day of midnight');
The string '+1 month first day of midnight' does three things.
'+1 month' - add one month.
'first day of' - switch to the first day of the month
'midnight' - change the time to 00:00:00
The modify method is really useful for many date manipulations!
So to answer the op, this is what you need — with a slight adjustment to format to cater for single digit months:
$expiryMonth = 6;
$expiryYear = 20;
$timezone = new DateTimeZone('Europe/London');
$expiryTime = \DateTime::createFromFormat(
'm-y',
$expiryMonth.'-'.$expiryYear,
$timezone
)->modify('+1 month first day of midnight');
$currentTime = new \DateTime('now', $timezone);
if ($expiryTime < $currentTime) {
// Card has expired.
}
An addition to the above answers.
Be aware that by default the days will also be in the calculation.
For example today is 2019-10-31 and if you run this:
\DateTime::createFromFormat('Ym', '202111');
It will output 2021-12-01, because day 31 does not exist in November and it will add 1 extra day to your DateTime object with a side effect that you will be in the month December instead of the expected November.
My suggestion is always use the day in your code.
For op's question:
$y=15;
$m=05;
if(strtotime( substr(date('Y'), 0, 2)."{$y}-{$m}" ) < strtotime( date("Y-m") ))
{
echo 'card is expired';
}
For others with full year:
$y=2015;
$m=5;
if(strtotime("{$y}-{$m}") < strtotime( date("Y-m") ))
{
echo 'card is expired';
}
Would it not be simpler to just compare the string "201709" to the current year-month? Creating datetime objects will cost php some effort, I suppose.
if($_POST['expYear']. str_pad($_POST['expMonth'],2,'0', STR_PAD_LEFT ) < date('Ym')) {
echo 'expired';
}
edited as Adam states
The best answer is provided by John Conde above. It it does the minimum amount of processing: creates two correct DateTime objects, compares them and that's all it needs.
It could work also as you started but you must format the dates in a way that puts the year first.
Think a bit about it: as dates, 08/15 (August 2015) is after 12/14 (December 2014) but as strings, '08 15' is before '12 14'.
When the year is in front, even as strings the years are compared first and then, only when the years are equal the months are compared:
$expiryDate = $expiryDateTime->format('y m');
$currentDate = $currentDateTime->format('y m');
if ($expiryDate < $currentDate) {
echo 'Expired';
} else {
echo 'Valid';
}
Keep it simple, as the answer above me says except you need to string pad to the left:
isCardExpired($month, $year)
{
$expires = $year.str_pad($month, 2, '0', STR_PAD_LEFT);
$now = date('Ym');
return $expires < $now;
}
No need to add extra PHP load using DateTime
If you are using Carbon, which is a very popular Datetime extension library. Then this should be:
$expMonth = $_POST['month'];
$expYear = $_POST['year'];
$format_m_y = str_pad($expMonth,2,'0', STR_PAD_LEFT).'-'.substr($expYear, 2);
$date = \Carbon\Carbon::createFromFormat('m-y', $format_m_y)
->endOfMonth()
->startOfDay();
if ($date->isPast()) {
// this card is expired
}
Also take into consideration the exact date and time expiration:
Credit cards expire at the end of the month printed as its expiration date, not at the beginning. Many cards actually technically expire one day after the end of that month. In any case, unless they list a specific day of expiration along with month and year, they should work all the way through the end of their expiration month. Cardholders should not wait until the last moment to secure a replacement card. Source

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