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');
Related
I have a following code, which would returns me previous and next date (selection by months). But I found it not working properly, when user selects a february. This is my code, which is applied to get the dates:
$dateSelect = #$_GET['datum'] ?: date('Y-m');
$date = DateTime::createFromFormat('Y-m', $dateSelect);
$prevdate = clone $date;
$date->modify('first day of next month');
$this->view->next_month = $date->format('m');
$this->view->next_year = $date->format('Y');
$prevdate->modify('first day of previous month');
$this->view->prev_month = $prevdate->format('m');
$this->view->prev_year = $prevdate->format('Y');
I have made an example so you can test it faster: PHP Sandbox
If user selects date 2017-02 then it is processed as 2017-03, which is incorrect. Try changing the value to 2017-01 in the example and you will see it is working properly. Since buttons in my code are filling with these values, my users can't go before February 2017 without manually changing the URL.
As DateTime will populate the missing d part of the date based on the current date, you're effectively getting 2017-02-29 (given that today is the 29th), which isn't valid, but DateTime adjusts to a valid date by rolling forward into the following month, giving 2017-03-01.
You need to force a d part to the date to avoid this default behaviour
$date = DateTime::createFromFormat('Y-m-d', $dateSelect . '-01');
Suppose I have a date specific as 2016-11-14 .
Also I have another date (starting) 02-14-2017 and ending 02-30-2017
I want to count the months that has passed is the date 2016-11-14 until the set of Range date.
I tried searching but I cant find a link on how to find months that has passed from a date to set of range date
DateTime() makes doing date math easy:
$datetime1 = new DateTime('2016-12-01');
$datetime2 = new DateTime('2017-02-01');
$interval = $datetime2->diff($datetime1);
echo (($interval->format('%y') * 12) + $interval->format('%m'));
Demo
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 want to find a specific week number from the specific start date. For example $date is dragged from the database (i.e. 07/08/2011)
I want this to be the start date so it would be week 3 now from this date. This is the code i have so far but just shows the ISO version:
$date = strtotime("".$row['start_date']."");
$weekNumber = date("W", $date);
print $weekNumber;
I have googled for past two hours but cannot seem to find any thing that resolves this! any help would be great thanks!
Get the difference between now and the startdate, and then divide by seven days (7*86400 seconds)
<?php
$startdate = strtotime("".$row['start_date']."");
$enddate = time();
$time_passed = $enddate - $startdate;
// if the first day after startdate is in "Week 1" according to your count
$weekcount_1 = ceil ( $time_passed / (86400*7));
// if the first day after startdate is in "Week 0" according to your count
$weekcount_0 = floor ( $time_passed / (86400*7));
?>
You can drag week number from db directly along with main date.
For example,
"SELECT start_date, (WEEK(NOW()) - WEEK(start_date)) as desired_week as week from table";