I am trying to create a booking form where a user can select a booking time between 2 given times in 5 minute intervals. For example I want time slots between 10am and 12pm which would give me about 20 time slots.
When the user goes to select a slot, the earliest slot should be at least 15 mins ahead of the current time but the user can select a slot and hour or more if desired.
I found some code on SO (can't remember where) and I've edited it for my needs and it works if the current time is within the start and end time but if the current time is an hour before the earliest time, it doesn't create the time slots.
I know why it does it but i don't know how to fix it. It has to do with the while condition.
I would like to be able to book a slot hours before the first available slot if that is possible.
$timenow = time();
$start_time = strtotime('+15 minutes', $timenow);
// round to next 15 minutes (15 * 60 seconds)
$start_time = ceil($start_time / (5 * 60)) * (5 * 60);
//set the start times
$opentime = strtotime('10:00');
$closetime = strtotime('11:55');
// get a list of prebooked slots from database
$time_slots = $this->countStartTimes();
$available_slots = array();
while($start_time <= $closetime && $start_time >= $opentime) {
$key = date('H:i', $start_time);
if(array_key_exists($key, $time_slots)) {
if($time_slots[$key] == SLOTS) {
$available_slots[] = 'FULL';
break;
}
}
$available_slots[] = date('H:i', $start_time);
$start_time = strtotime('+5 minutes', $start_time);
}
I managed to get it working using Datetime()
$timenow = new DateTime(date('H:i'));
$timenow->add(new DateInterval('PT15M'));
$start = new DateTime('11:00');
$end = new DateTime('14:00');
$interval = new DateInterval('PT5M');
$time_slots = $this->countStartTimes();
$available_slots = array();
$period = new DatePeriod($start, $interval, $end);
foreach($period as $time) {
$timeslot = $time->format('H:i');
if ($timenow > $time) {
continue;
}
if(array_key_exists($timeslot, $time_slots)) {
if($time_slots[$timeslot] == SLOTS) {
$available_slots[] = array('key' => $timeslot, 'value' => 'FULL');
continue;
}
}
$available_slots[] = array('key' => $timeslot, 'value' => $timeslot);
}
Carbon has all of the functions inherited from the base DateTime class. This approach allows you to access the base functionality if you see anything missing in Carbon but is there in DateTime.
// Carbon::diffInYears(Carbon $dt = null, $abs = true)
echo Carbon::now('America/Vancouver')->diffInSeconds(Carbon::now('Europe/London')); // 0
$dtOttawa = Carbon::createFromDate(2000, 1, 1, 'America/Toronto');
$dtVancouver = Carbon::createFromDate(2000, 1, 1, 'America/Vancouver');
echo $dtOttawa->diffInHours($dtVancouver); // 3
echo $dtOttawa->diffInHours($dtVancouver, false); // 3
echo $dtVancouver->diffInHours($dtOttawa, false);
Use carbon class for this it really help you
Related
I tried #ebelendez's code for Calculating working hours between two dates, however I'm confused on how to set the value of Saturdays by 3 hours (08:00-11:00). For example, the working hours per day during weekdays is 8 hours (excluding 1 hour break), let's say I want to get the total working hours from Thursday to Saturday, the expected result would be 19 hours.
Here is what I've done. Can someone help me with this?
$from = '2022-04-21 07:00:00';
$to = '2022-04-23 16:00:00';
echo abs(get_working_hours($from, $to));
function get_working_hours($from,$to){
//config
$ini_time = [7,0]; //hr, min
$end_time = [16,0]; //hr, min
//date objects
$ini = date_create($from);
$ini_wk = date_time_set(date_create($from),$ini_time[0],$ini_time[1]);
$end = date_create($to);
$end_wk = date_time_set(date_create($to),$end_time[0],$end_time[1]);
//days
$workdays_arr = get_workdays($ini,$end);
$workdays_count = count($workdays_arr);
$workday_seconds = (($end_time[0] * 60 + $end_time[1]) - ($ini_time[0] * 60 + $ini_time[1])) * 60 - 3600;
//get time difference
$ini_seconds = 0;
$end_seconds = 0;
if(in_array($ini->format('Y-m-d'),$workdays_arr)) $ini_seconds = $ini->format('U') - $ini_wk->format('U');
if(in_array($end->format('Y-m-d'),$workdays_arr)) $end_seconds = $end_wk->format('U') - $end->format('U');
$seconds_dif = $ini_seconds > 0 ? $ini_seconds : 0;
if($end_seconds > 0) $seconds_dif += $end_seconds;
//final calculations
$working_seconds = ($workdays_count * $workday_seconds) - $seconds_dif;
return $working_seconds / 3600; //return hrs
}
function get_workdays($ini,$end){
//config
$skipdays = [0]; //sunday:0
$skipdates = [];
//vars
$current = clone $ini;
$current_disp = $current->format('Y-m-d');
$end_disp = $end->format('Y-m-d');
$days_arr = [];
//days range
while($current_disp <= $end_disp){
if(!in_array($current->format('w'),$skipdays) && !in_array($current_disp,$skipdates)){
$days_arr[] = $current_disp;
}
$current->add(new DateInterval('P1D')); //adds one day
$current_disp = $current->format('Y-m-d');
}
return $days_arr;
}
Your code and linked answers seem unnecessarily complicated. All we really need is to:
Configure how many hours should be counted for for each day;
Create an iterable DatePeriod (with DateTime objects for each date in the period);
Iterate dates, look up how many hours should be counted for each day, sum it up.
class CountWorkingHours
{
// Define hours counted for each day:
public array $hours = [
'Mon' => 8,
'Tue' => 8,
'Wed' => 8,
'Thu' => 8,
'Fri' => 8,
'Sat' => 3,
'Sun' => 0
];
// Method for counting the hours:
public function get_hours_for_period(string $from, string $to): int
{
// Create DatePeriod with requested Start/End dates:
$period = new DatePeriod(
new DateTime($from),
new DateInterval('P1D'),
new DateTime($to)
);
$hours = [];
// Loop over DateTime objects in the DatePeriod:
foreach($period as $date) {
// Get name of day and add matching hours:
$day = $date->format('D');
$hours[] = $this->hours[$day];
}
// Return sum of hours:
return array_sum($hours);
}
}
Source # BitBucket
Usage (returns an integer with working hours in a given period):
$cwh = new CountWorkingHours();
$hours = $cwh->get_hours_for_period('2022-04-21 07:00:00', '2022-04-30 16:00:00');
// = 62
If you need to account for public holidays etc. exceptions to the standard weekly hour counts, you can add a check inside the period loop for "skip dates". For example, have a $skip_dates property with an array of non-work-days, then check for !in_array($date->format('Y-m-d'), $this->skip_dates) before incrementing the work hours for a given day.
P.S. This code assumes that you are calculating whole working days. If your start or end hours were defined in the middle of a working day, that wouldn't be accounted for. (If you wanted to factor that in, you'd have to configure daily work times and the code would have to account for that in evaluating start and end dates. Seemed an unnecessary exercise for current purposes.)
I have to create a scheduling component that will plan e-mails that need to be sent out. Users can select a start time, end time, and frequency. Code should produce a random moment for every frequency, between start and end time. Outside of office hours.
Paramaters:
User can select a period between 01/01/2020 (the start) and 01/01/2021 (the end). In this case user selects a timespan of one exactly year.
User can select a frequency. In this case user selects '2 months'.
Function:
Code produces a list of datetimes. The total time (one year) is divided by frequency (2 months). We expect a list of 6 datetimes.
Every datetime is a random moment in said frequency (2 months). Within office hours.
Result:
An example result for these paramaters might as follows, with the calculated frequency bounds for clarity:
[jan/feb] 21-02-2020 11.36
[mrt/apr] 04-03-2020 16.11
[mei/jun] 13-05-2020 09.49
[jul-aug] 14-07-2020 15.25
[sep-okt] 02-09-2020 14.09
[nov-dec] 25-12-2020 13.55
--
I've been thinking about how to implement this best, but I can't figure out an elegant solution.
How could one do this using PHP?
Any insights, references, or code spikes would be greatly appreciated. I'm really stuck on this one.
I think you're just asking for suggestions on how to generate a list of repeating (2 weekly) dates with a random time between say 9am and 5pm? Is that right?
If so - something like this (untested, pseudo code) might be a starting point:
$start = new Datetime('1st January 2021');
$end = new Datetime('1st July 2021');
$day_start = 9;
$day_end = 17;
$date = $start;
$dates = [$date]; // Start date into array
while($date < $end) {
$new_date = clone($date->modify("+ 2 weeks"));
$new_date->setTime(mt_rand($day_start, $day_end), mt_rand(0, 59));
$dates[] = $new_date;
}
var_dump($dates);
Steve's anwser seems good, but you should consider 2 additional things
holiday check, in the while after first $new_date line, like:
$holiday = array('2021-01-01', '2021-01-06', '2021-12-25');
if (!in_array($new_date,$holiday))
also a check if date is a office day or a weekend in a similar way as above with working days as an array.
It's kind of crappy code but I think it will work as you wish.
function getDiffInSeconds(\DateTime $start, \DateTime $end) : int
{
$startTimestamp = $start->getTimestamp();
$endTimestamp = $end->getTimestamp();
return $endTimestamp - $startTimestamp;
}
function getShiftData(\DateTime $start, \DateTime $end) : array
{
$shiftStartHour = \DateTime::createFromFormat('H:i:s', $start->format('H:i:s'));
$shiftEndHour = \DateTime::createFromFormat('H:i:s', $end->format('H:i:s'));
$shiftInSeconds = intval($shiftEndHour->getTimestamp() - $shiftStartHour->getTimestamp());
return [
$shiftStartHour,
$shiftEndHour,
$shiftInSeconds,
];
}
function dayIsWeekendOrHoliday(\DateTime $date, array $holidays = []) : bool
{
$weekendDayIndexes = [
0 => 'Sunday',
6 => 'Saturday',
];
$dayOfWeek = $date->format('w');
if (empty($holidays)) {
$dayIsWeekendOrHoliday = isset($weekendDayIndexes[$dayOfWeek]);
} else {
$dayMonthDate = $date->format('d/m');
$dayMonthYearDate = $date->format('d/m/Y');
$dayIsWeekendOrHoliday = (isset($weekendDayIndexes[$dayOfWeek]) || isset($holidays[$dayMonthDate]) || isset($holidays[$dayMonthYearDate]));
}
return $dayIsWeekendOrHoliday;
}
function getScheduleDates(\DateTime $start, \DateTime $end, int $frequencyInSeconds) : array
{
if ($frequencyInSeconds < (24 * 60 * 60)) {
throw new \InvalidArgumentException('Frequency must be bigger than one day');
}
$diffInSeconds = getDiffInSeconds($start, $end);
// If difference between $start and $end is bigger than two days
if ($diffInSeconds > (2 * 24 * 60 * 60)) {
// If difference is bigger than 2 days we add 1 day to start and subtract 1 day from end
$start->modify('+1 day');
$end->modify('-1 day');
// Getting new $diffInSeconds after $start and $end changes
$diffInSeconds = getDiffInSeconds($start, $end);
}
if ($frequencyInSeconds > $diffInSeconds) {
throw new \InvalidArgumentException('Frequency is bigger than difference between dates');
}
$holidays = [
'01/01' => 'New Year',
'18/04/2020' => 'Easter 1st official holiday because 19/04/2020',
'20/04/2020' => 'Easter',
'21/04/2020' => 'Easter 2nd day',
'27/04' => 'Konings',
'04/05' => '4mei',
'05/05' => '4mei',
'24/12' => 'Christmas 1st day',
'25/12' => 'Christmas 2nd day',
'26/12' => 'Christmas 3nd day',
'27/12' => 'Christmas 3rd day',
'31/12' => 'Old Year'
];
[$shiftStartHour, $shiftEndHour, $shiftInSeconds] = getShiftData($start, $end);
$amountOfNotifications = floor($diffInSeconds / $frequencyInSeconds);
$periodInSeconds = intval($diffInSeconds / $amountOfNotifications);
$maxDaysBetweenNotifications = intval($periodInSeconds / (24 * 60 * 60));
// If $maxDaysBetweenNotifications is equals to 1 then we have to change $periodInSeconds to amount of seconds for one day
if ($maxDaysBetweenNotifications === 1) {
$periodInSeconds = (24 * 60 * 60);
}
$dates = [];
for ($i = 0; $i < $amountOfNotifications; $i++) {
$periodStart = clone $start;
$periodStart->setTimestamp($start->getTimestamp() + ($i * $periodInSeconds));
$seconds = mt_rand(0, $shiftInSeconds);
// If $maxDaysBetweenNotifications is equals to 1 then we have to check only one day without loop through the dates
if ($maxDaysBetweenNotifications === 1) {
$interval = new \DateInterval('P' . $maxDaysBetweenNotifications . 'DT' . $seconds . 'S');
$date = clone $periodStart;
$date->add($interval);
$dayIsWeekendOrHoliday = dayIsWeekendOrHoliday($date, $holidays);
} else {
// When $maxDaysBetweenNotifications we have to loop through the dates to pick them
$loopsCount = 0;
$maxLoops = 3; // Max loops before breaking and skipping the period
do {
$day = mt_rand(0, $maxDaysBetweenNotifications);
$periodStart->modify($shiftStartHour);
$interval = new \DateInterval('P' . $day . 'DT' . $seconds . 'S');
$date = clone $periodStart;
$date->add($interval);
$dayIsWeekendOrHoliday = dayIsWeekendOrHoliday($date, $holidays);
// If the day is weekend or holiday then we have to increment $loopsCount by 1 for each loop
if ($dayIsWeekendOrHoliday === true) {
$loopsCount++;
// If $loopsCount is equals to $maxLoops then we have to break the loop
if ($loopsCount === $maxLoops) {
break;
}
}
} while ($dayIsWeekendOrHoliday);
}
// Adds the date to $dates only if the day is not a weekend day and holiday
if ($dayIsWeekendOrHoliday === false) {
$dates[] = $date;
}
}
return $dates;
}
$start = new \DateTime('2020-12-30 08:00:00', new \DateTimeZone('Europe/Sofia'));
$end = new \DateTime('2021-01-18 17:00:00', new \DateTimeZone('Europe/Sofia'));
$frequencyInSeconds = 86400; // 1 day
$dates = getScheduleDates($start, $end, $frequencyInSeconds);
var_dump($dates);
You have to pass $start, $end and $frequencyInSeconds as I showed in example and then you will get your random dates. Notice that I $start and $end must have hours in them because they are used as start and end hours for shifts. Because the rule is to return a date within a shift time only in working days. Also you have to provide frequency in seconds - you can calculate them outside the function or you can change it to calculate them inside. I did it this way because I don't know what are your predefined periods.
This function returns an array of \DateTime() instances so you can do whatever you want with them.
UPDATE 08/01/2020:
Holidays now are part of calculation and they will be excluded from returned dates if they are passed when you are calling the function. You can pass them in d/m and d/m/Y formats because of holidays like Easter and in case when the holiday is on weekend but people will get additional dayoff during the working week.
UPDATE 13/01/2020:
I've made updated code version to fix the issue with infinite loops when $frequencyInSeconds is shorter like 1 day. The new code used few functions getDiffInSeconds, getShiftData and dayIsWeekendOrHoliday as helper methods to reduce code duplication and cleaner and more readable code
I have a function to return the difference between 2 dates, however I need to work out the difference in working hours, assuming Monday to Friday (9am to 5:30pm):
//DATE DIFF FUNCTION
// Set timezone
date_default_timezone_set("GMT");
// Time format is UNIX timestamp or
// PHP strtotime compatible strings
function dateDiff($time1, $time2, $precision = 6) {
// If not numeric then convert texts to unix timestamps
if (!is_int($time1)) {
$time1 = strtotime($time1);
}
if (!is_int($time2)) {
$time2 = strtotime($time2);
}
// If time1 is bigger than time2
// Then swap time1 and time2
if ($time1 > $time2) {
$ttime = $time1;
$time1 = $time2;
$time2 = $ttime;
}
// Set up intervals and diffs arrays
$intervals = array('year','month','day','hour','minute','second');
$diffs = array();
// Loop thru all intervals
foreach ($intervals as $interval) {
// Set default diff to 0
$diffs[$interval] = 0;
// Create temp time from time1 and interval
$ttime = strtotime("+1 " . $interval, $time1);
// Loop until temp time is smaller than time2
while ($time2 >= $ttime) {
$time1 = $ttime;
$diffs[$interval]++;
// Create new temp time from time1 and interval
$ttime = strtotime("+1 " . $interval, $time1);
}
}
$count = 0;
$times = array();
// Loop thru all diffs
foreach ($diffs as $interval => $value) {
// Break if we have needed precission
if ($count >= $precision) {
break;
}
// Add value and interval
// if value is bigger than 0
if ($value > 0) {
// Add s if value is not 1
if ($value != 1) {
$interval .= "s";
}
// Add value and interval to times array
$times[] = $value . " " . $interval;
$count++;
}
}
// Return string with times
return implode(", ", $times);
}
Date 1 = 2012-03-24 03:58:58
Date 2 = 2012-03-22 11:29:16
Is there a simple way of doing this, i.e - calculating the percentage of working hours in a week and dividing the difference using the above function - I have played around with this idea and got some very strange figures...
Or is there better way....?
This example uses PHP's built in DateTime classes to do the date math. How I approached this was to start by counting the number of full working days between the two dates and then multiply that by 8 (see notes). Then it gets the hours worked on the partial days and adds them to the total hours worked. Turning this into a function would be fairly straightforward to do.
Notes:
Does not take timestamps into account. But you already know how to do that.
Does not handle holidays. (That can be easily added by using an array of holidays and adding it to where you filter out Saturdays and Sundays).
Requires PHP 5.3.6+
Assumes an 8 hour workday. If employees do not take lunch change $hours = $days * 8; to $hours = $days * 8.5;
.
<?php
// Initial datetimes
$date1 = new DateTime('2012-03-22 11:29:16');
$date2 = new DateTime('2012-03-24 03:58:58');
// Set first datetime to midnight of next day
$start = clone $date1;
$start->modify('+1 day');
$start->modify('midnight');
// Set second datetime to midnight of that day
$end = clone $date2;
$end->modify('midnight');
// Count the number of full days between both dates
$days = 0;
// Loop through each day between two dates
$interval = new DateInterval('P1D');
$period = new DatePeriod($start, $interval, $end);
foreach ($period as $dt) {
// If it is a weekend don't count it
if (!in_array($dt->format('l'), array('Saturday', 'Sunday'))) {
$days++;
}
}
// Assume 8 hour workdays
$hours = $days * 8;
// Get the number of hours worked on the first day
$date1->modify('5:30 PM');
$diff = $date1->diff($start);
$hours += $diff->h;
// Get the number of hours worked the second day
$date1->modify('8 AM');
$diff = $date2->diff($end);
$hours += $diff->h;
echo $hours;
See it in action
Reference
DateTime Class
DatePeriod Class
DateInterval Class
Here's what I've come up with.
My solution checks the start and end times of the original dates, and adjusts them according to the actual start and end times of the work day (if the original start time is before work's opening time, it sets it to the latter).
After this is done to both start and end times, the times are compared to retrieve a DateInterval diff, calculating the total days, hours, etc. The date range is then checked for any weekend days, and if found, one total day is reduced from the diff.
Finally, the hours are calculated as commented. :)
Cheers to John for inspiring some of this solution, particularly the DatePeriod to check for weekends.
Gold star to anyone who breaks this; I'll be happy to update if anyone finds a loophole!
Gold star to myself, I broke it! Yeah, weekends are still buggy (try starting at 4pm on Saturday and ending at 1pm Monday). I will conquer you, work hours problem!
Ninja edit #2: I think I took care of the weekend bugs by reverting the start and end times to the most recent respective weekday if they fall on a weekend. Got good results after testing a handful of date ranges (starting and ending on the same weekend barfs, as expected). I'm not entirely convinced this is as optimized / simple as it could be, but at least it works better now.
// Settings
$workStartHour = 9;
$workStartMin = 0;
$workEndHour = 17;
$workEndMin = 30;
$workdayHours = 8.5;
$weekends = ['Saturday', 'Sunday'];
$hours = 0;
// Original start and end times, and their clones that we'll modify.
$originalStart = new DateTime('2012-03-22 11:29:16');
$start = clone $originalStart;
// Starting on a weekend? Skip to a weekday.
while (in_array($start->format('l'), $weekends))
{
$start->modify('midnight tomorrow');
}
$originalEnd = new DateTime('2012-03-24 03:58:58');
$end = clone $originalEnd;
// Ending on a weekend? Go back to a weekday.
while (in_array($end->format('l'), $weekends))
{
$end->modify('-1 day')->setTime(23, 59);
}
// Is the start date after the end date? Might happen if start and end
// are on the same weekend (whoops).
if ($start > $end) throw new Exception('Start date is AFTER end date!');
// Are the times outside of normal work hours? If so, adjust.
$startAdj = clone $start;
if ($start < $startAdj->setTime($workStartHour, $workStartMin))
{
// Start is earlier; adjust to real start time.
$start = $startAdj;
}
else if ($start > $startAdj->setTime($workEndHour, $workEndMin))
{
// Start is after close of that day, move to tomorrow.
$start = $startAdj->setTime($workStartHour, $workStartMin)->modify('+1 day');
}
$endAdj = clone $end;
if ($end > $endAdj->setTime($workEndHour, $workEndMin))
{
// End is after; adjust to real end time.
$end = $endAdj;
}
else if ($end < $endAdj->setTime($workStartHour, $workStartMin))
{
// End is before start of that day, move to day before.
$end = $endAdj->setTime($workEndHour, $workEndMin)->modify('-1 day');
}
// Calculate the difference between our modified days.
$diff = $start->diff($end);
// Go through each day using the original values, so we can check for weekends.
$period = new DatePeriod($start, new DateInterval('P1D'), $end);
foreach ($period as $day)
{
// If it's a weekend day, take it out of our total days in the diff.
if (in_array($day->format('l'), ['Saturday', 'Sunday'])) $diff->d--;
}
// Calculate! Days * Hours in a day + hours + minutes converted to hours.
$hours = ($diff->d * $workdayHours) + $diff->h + round($diff->i / 60, 2);
As the old saying goes "if you want something done right do it yourself". Not saying this is optimal but its atleast returning the correct amount of hours for me.
function biss_hours($start, $end){
$startDate = new DateTime($start);
$endDate = new DateTime($end);
$periodInterval = new DateInterval( "PT1H" );
$period = new DatePeriod( $startDate, $periodInterval, $endDate );
$count = 0;
foreach($period as $date){
$startofday = clone $date;
$startofday->setTime(8,30);
$endofday = clone $date;
$endofday->setTime(17,30);
if($date > $startofday && $date <= $endofday && !in_array($date->format('l'), array('Sunday','Saturday'))){
$count++;
}
}
//Get seconds of Start time
$start_d = date("Y-m-d H:00:00", strtotime($start));
$start_d_seconds = strtotime($start_d);
$start_t_seconds = strtotime($start);
$start_seconds = $start_t_seconds - $start_d_seconds;
//Get seconds of End time
$end_d = date("Y-m-d H:00:00", strtotime($end));
$end_d_seconds = strtotime($end_d);
$end_t_seconds = strtotime($end);
$end_seconds = $end_t_seconds - $end_d_seconds;
$diff = $end_seconds-$start_seconds;
if($diff!=0):
$count--;
endif;
$total_min_sec = date('i:s',$diff);
return $count .":".$total_min_sec;
}
$start = '2014-06-23 12:30:00';
$end = '2014-06-27 15:45:00';
$go = biss_hours($start,$end);
echo $go;
I want to calculate all Sunday's between given two dates. I tried following code. It works fine if days are less but if i enter more days. It keeps processing and Maximum execution time exceeds i changed the time but it even keeps processing even execution time is 200sec.
code is
<?php
$one="2013-01-01";
$two="2013-02-30";
$no=0;
for($i=$one;$i<=$two;$i++)
{
$day=date("N",strtotime($i));
if($day==7)
{
$no++;
}
}
echo $no;
?>
please help.
John Conde's answer is correct, but here is a more efficient and mathy solution:
$start = new DateTime('2013-01-06');
$end = new DateTime('2013-01-20');
$days = $start->diff($end, true)->days;
$sundays = intval($days / 7) + ($start->format('N') + $days % 7 >= 7);
echo $sundays;
Let me break it down for you.
$start = new DateTime('2013-01-06');
$end = new DateTime('2013-01-20');
First, create some DateTime objects, which are powerful built-in PHP objects meant for exactly this kind of problem.
$days = $start->diff($end, true)->days;
Next, use DateTime::diff to find the difference from $start to $end (passing true here as the second parameter ensures that this value is always positive), and get the number of days between them.
$sundays = intval($days / 7) + ($start->format('N') + $days % 7 >= 7);
Here comes the big one - but it's not so complicated, really. First, we know there is one Sunday for every week, so we have at least $days / 7 Sundays to begin with, rounded down to the nearest int with intval.
On top of that, there could be a Sunday in a span of time less than a week; for example, Friday to Monday of the next week contains 4 days; one of them is a Sunday. So, depending on when we start and end, there could be another. This is easy to account for:
$start->format('N') (see DateTime::format) gives us the ISO-8601 day of the week for the start date, which is a number from 1 to 7 (1 is Monday, 7 is Sunday).
$days % 7 gives us the number of leftover days that don't divide evenly into weeks.
If our starting day and the number of leftover days add up to 7 or more, then we reached a Sunday. Knowing that, we just have to add that expression, which will give us 1 if it's true or 0 if it's false, since we're adding it to an int value.
And there you have it! The advantage of this method is that it doesn't require iterating over every day between the given times and checking to see if it's a Sunday, which will save you a lot computation, and also it will make you look really clever. Hope that helps!
<?php
$no = 0;
$start = new DateTime('2013-01-01');
$end = new DateTime('2013-04-30');
$interval = DateInterval::createFromDateString('1 day');
$period = new DatePeriod($start, $interval, $end);
foreach ($period as $dt)
{
if ($dt->format('N') == 7)
{
$no++;
}
}
echo $no;
See it in action
Here is a solution if you want the Sundays in a specific date range.
function dateRange($begin, $end, $interval = null)
{
$begin = new DateTime($begin);
$end = new DateTime($end);
$end = $end->modify('+1 day');
$interval = new DateInterval($interval ? $interval : 'P1D');
return iterator_to_array(new DatePeriod($begin, $interval, $end));
}
/* define date range */
$dates = dateRange('2018-03-01', '2018-03-31');
/* define weekdays */
$weekends = array_filter($dates, function ($date) {
$day = $date->format("N");
return $day === '6' || $day === '7';
});
/* weekdays output */
foreach ($weekends as $date) {
echo $date->format("D Y-m-d") . "</br>";
}
/* define sundays */
$sundays = array_filter($dates, function ($date) {
return $date->format("N") === '7';
});
/* sundays output */
foreach ($sundays as $date) {
echo $date->format("D Y-m-d") . "</br>";
}
/* define mondays */
$mondays = array_filter($dates, function ($date) {
return $date->format("N") === '1';
});
/* mondays output */
foreach ($mondays as $date) {
echo $date->format("D Y-m-d") . "</br>";
}
Just change the number for any days you want in your output:
Monday = 1
Tuesday = 2
Wednesday = 3
Thursday = 4
Friday = 5
Saturday = 6
Sunday = 7
I'm trying to work with dates for the first time, I did it something about that with Flash but it's different.
I have two different dates and I'd like to see the difference in hours and days with them, I've found too many examples but not what I'm loking for:
<?php
$now_date = strtotime (date ('Y-m-d H:i:s')); // the current date
$key_date = strtotime (date ("2009-11-21 14:08:42"));
print date ($now_date - $key_date);
// it returns an integer like 5813, 5814, 5815, etc... (I presume they are seconds)
?>
How can I convert it to hours or to days?
The DateTime diff function returns a DateInterval object. This object consists of variabeles related to the difference. You can query the days, hours, minutes, seconds just like in the example above.
Example:
<?php
$dateObject = new DateTime(); // No arguments means 'now'
$otherDateObject = new DateTime('2008-08-14 03:14:15');
$diffObject = $dateObject->diff($otherDateObject));
echo "Days of difference: ". $diffObject->days;
?>
See the manual about DateTime.
Sadly, it's a PHP 5.3> only feature.
Well, you can always use date_diff, but that is only for PHP 5.3.0+
The alternative would be math.
How can I convert it [seconds] to hours or to days?
There are 60 seconds per minute, which means there are 3600 seconds per hour.
$hours = $seconds/3600;
And, of course, if you need days ...
$days = $hours/24;
If you dont have PHP5.3 you could use this method from userland (taken from WebDeveloper.com)
function date_time_diff($start, $end, $date_only = true) // $start and $end as timestamps
{
if ($start < $end) {
list($end, $start) = array($start, $end);
}
$result = array('years' => 0, 'months' => 0, 'days' => 0);
if (!$date_only) {
$result = array_merge($result, array('hours' => 0, 'minutes' => 0, 'seconds' => 0));
}
foreach ($result as $period => $value) {
while (($start = strtotime('-1 ' . $period, $start)) >= $end) {
$result[$period]++;
}
$start = strtotime('+1 ' . $period, $start);
}
return $result;
}
$date_1 = strtotime('2005-07-31');
$date_2 = time();
$diff = date_time_diff($date_1, $date_2);
foreach ($diff as $key => $val) {
echo $val . ' ' . $key . ' ';
}
// Displays:
// 3 years 4 months 11 days
TheGrandWazoo mentioned a method for php 5.3>. For lower versions you can devide the number of seconds between the two dates with the number of seconds in a day to find the number of days.
For days, you do:
$days = floor(($now_date - $key_date) / (60 * 60 * 24))
If you want to know how many hours are still left, you can use the modulo operator (%)
$hours = floor((($now_date - $key_date) % * (60 * 60 * 24)) / 60 * 60)
<?php
$now_date = strtotime (date ('Y-m-d H:i:s')); // the current date
$key_date = strtotime (date ("2009-11-21 14:08:42"));
$diff = $now_date - $key_date;
$days = floor($diff/(60*60*24));
$hours = floor(($diff-($days*60*60*24))/(60*60));
print $days." ".$hours." difference";
?>
I prefer to use epoch/unix time deltas. Time represented in seconds and as such you can very quickly divide by 3600 for hours and divide by 24*3600=86400 for days.