I have written a round robin tournament generator in PHP for an online electronic sports league, and I need to calculate the dates for each game in the tournament. Games are played every Thursday and Sunday over the course of many weeks (the number of weeks is dependent on how many teams are participating). Given the starting week and the number of weeks what would be the best way to calculate those dates?
I'm guessing it requires using some combination of DateTime, DateInterval, and DatePeriod; but I am having trouble figuring out how it would be done.
Update:
Apologies for not providing the code before. Here is the solution I had originally come up with. I didn't know if there was a simpler way of doing it. The function was called submitSchedule where the dates were generated.
<html>
<body>
<?php
function roundRobin($teams) {
$len = count($teams);
$schedule = array();
for ($i = 0; $i < $len - 1; $i++)
{
$home = array_slice($teams, 0, $len / 2);
$away = array_slice($teams, $len / 2);
$day = array();
for ($j = 0; $j < $len / 2; $j++)
{
array_push($day, array($home[$j], $away[$j]));
}
array_push($schedule, $day);
$temp = $away[0];
for ($j = 0; $j < count($away) - 1; $j++)
{
$away[$j] = $away[$j + 1];
}
$away[count($away) - 1] = $home[count($home) - 1];
for ($j = count($home) - 1; $j > 1; $j--)
{
$home[$j] = $home[$j - 1];
}
$home[1] = $temp;
$teams = array_merge($home, $away);
}
return $schedule;
}
function roundRobinBalanced($teams)
{
$schedule = roundRobin($teams);
for ($i = 1; $i < count($schedule); $i+=2)
{
$schedule[$i][0] = array_reverse($schedule[$i][0]);
}
return $schedule;
}
function doubleRoundRobinBalanced($teams)
{
$sched2 = roundRobinBalanced($teams);
for ($i = 0; $i < count($sched2); $i++)
{
$sched2[$i] = array_reverse($sched2[$i]);
}
return array_merge(roundRobinBalanced($teams), $sched2);
}
function tripleRoundRobinBalanced($teams)
{
return array_merge(doubleRoundRobinBalanced($teams), roundRobinBalanced($teams));
}
function submitSchedule($schedule, $start, $intervals, &$con)
{
mysqli_query($con, "TRUNCATE TABLE matches");
$curDate = $start;
echo "<pre>";
for ($i = 0; $i < count($schedule); $i++)
{
for ($j = 0; $j < count($schedule[$i]); $j++)
{
$temp0 = $schedule[$i][$j][0];
$temp1 = $schedule[$i][$j][1];
$temp2 = date_format($curDate, "Y-m-d");
mysqli_query($con,"INSERT INTO matches (T1ID, T2ID, gameDate) VALUES ('$temp0', '$temp1', '$temp2')");
echo "<span style=\"background:lightblue;\">( " . date_format(new DateTime(), "Y-m-d H:i:s") . " )</span>" . "> INSERT INTO matches (T1ID, T2ID, gameDate) VALUES (". $schedule[$i][$j][0] . ", " . $schedule[$i][$j][1] . ", \"" . date_format($curDate, "Y-m-d") . "\")<br>";
}
date_add($curDate, date_interval_create_from_date_string($intervals[$i % count($intervals)]));
}
echo "</pre>";
}
$teams = array();
$con=mysqli_connect("localhost:3306","root","REMOVED","schedule");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
//Select all items from the 'teams' table and order them first in descending order by points, then in ascending order by 'teamName'
$result = mysqli_query($con,"SELECT * FROM teams");
while($row = mysqli_fetch_array($result))
{
array_push($teams, $row['TID']);
}
if (count($teams) % 2 == 1)
{
array_push($teams, null);
}
shuffle($teams);
$schedule = tripleRoundRobinBalanced($teams);
// echo "<pre>" . json_encode($schedule, JSON_PRETTY_PRINT, JSON_FORCE_OBJECT) . "</pre>";
echo "<pre>";
print_r($schedule);
echo "</pre>";
// ---- List of possible DateTime expressions ----
// thursday
// next thursday
// YYYY-MM-DD
// DD/MM/yy
// thursday + 1 day
$start = new DateTime("thursday"); // Indicates the date of the first game
$jump = array("3 days", "4 days"); // Indicates the time intervals of each game (e.g. If day 1 starts on thursday, day 2 starts on sunday, day 3 starts on thursday, etc.)
submitSchedule($schedule, $start, $jump, $con);
mysqli_close($con);
?>
</body>
</html>
The way to achieve this is by using PHP's DateTime classes as you guessed. They are really very useful. I would suggest a little function something like this:-
/**
* #param $startWeek ISO week number of the first week
* #param $numWeeks The number of weeks to run including the first
*
* #return \DateTime[] An array of DateTime objects
*/
function getPlayDays($startWeek, $numWeeks)
{
$numWeeks --;
$result = [];
$currYear = (int)(new \DateTime())->format('Y');
$oneDay = new \DateInterval('P1D');
// Start on the first Thursday of the given week.
$startDate = (new \DateTime())->setISODate($currYear, $startWeek, 4);
$endDate = clone $startDate;
$endDate->add(new \DateInterval("P{$numWeeks}W"));
// End on the Sunday of the last week.
$endDate->setISODate((int)$endDate->format('o'), (int)$endDate->format('W'), 7);
$period = new \DatePeriod($startDate, $oneDay, $endDate->add($oneDay));
foreach($period as $day){
if(4 === (int)$day->format('N') || 7 === (int)$day->format('N') ){
$result[] = $day;
}
}
return $result;
}
foreach(getPlayDays(1, 3) as $playDay){
var_dump($playDay->format('D m-d-Y'));
}
You didn't specify how you would identify the starting week, so I have assumed an ISO week number.
Output:-
string 'Thu 01-02-2014' (length=14)
string 'Sun 01-05-2014' (length=14)
string 'Thu 01-09-2014' (length=14)
string 'Sun 01-12-2014' (length=14)
string 'Thu 01-16-2014' (length=14)
string 'Sun 01-19-2014' (length=14)
See it working.
DateTime manual.
This function will quite happily cope with DST changes, leap years and weeks close to the start and end of the year thanks to the built in magic of the DateTime classes :)
proof or STFU.
Have a look at the strtotime function:
http://www.php.net/manual/en/function.strtotime.php
You could do something like this:
$startDate = "May 15, 2014";
$startDate = strtotime($startDate);
And you could get the start of the Sunday match by simply adding three days:
$nextDate = strtotime("+3 day", $startDate);
Your question is a bit vague, but I think this is what you were asking.
Let's say you have the timestamps of the day of starting week (06:00 AM time)...
Every other date will be 7 days (86400 seconds * 7) in the future.
Let's assume you will run this for 52 weeks (1 year)
$firstThu = 1234567890;
$firstSun = 9876543210;
$nextThus = array();
$nextSuns = array();
for($i=0; $i<52; $i++) {
$nextThus[] = date('d/m/Y', $firstThu + (86400 * 7 * $i));
$nextSuns[] = date('d/m/Y', $firstSun + (86400 * 7 * $i));
}
At the end of the loop you will have two arrays with all the 52 weeks dates.
i want to create a select field where you can select all calendar weeks between the last 5 and the next 5.
It is important, to notice, that some of the last 5 weeks are in 2013 and some in 2014.
My starting point is:
function generateSelect($calendarweek)
{
$this_cw = $calenderweek;
$return = "<select>";
list($y, $m, $d) = explode('-', date('Y-m-d'));
for ($i=-5; $i<=5; $i++)
{
$cw = date('W', mktime(0, 0, 0, $m, $d + 7*$i, $y));
$return .= "<option>".$cw." HERE I WANT THE CORRESPONDING YEAR</option>";
}
$return .= "</select>";
return $return;
}
My main problem is the corresponding year.
And I don't know if my solution is that good.
$calendarweek is the current calenderweek
The answer to this question is almost identical to this one:
$weeks = 5; # number of weeks before and after current week
$dt = new DateTime;
$dt->modify("-$weeks week");
for ($i = 0; $i <= $weeks * 2; $i++) {
echo $dt->format('W o'), "\n";
$dt->modify('+1 week');
}
demo
or just change date formating in your example from $cw = date('W', to $cw = date('W o',, where format o is ISO-8601 year number, read more here.
I have a calendar that I want to allow events to be repeated on a week day of the month. Some examples would be:
Repeat every 4th Tuesday of the month
Repeat every 2nd Friday of the month
And so on...
What I need is the ability to find out how many week days (for example Tuesday's) have passed this month so far.
I found some code that returns how many Monday's have passed.
$now=time() + 86400;
if (($dow = date('w', $now)) == 0) $dow = 7;
$begin = $now - (86400 * ($dow-1));
echo "Mondays: ".ceil(date('d', $begin) / 7)."<br/>";
This works well but how do I make it so that I can determine any week day? I cannot seem to get my head around the code to make this work.
strtotime is really useful for this kind of thing. Here are lists of the supported syntax. Using your example of repeat every 2nd Friday of the month I wrote the following simple snippet for you:
<?php
$noOfMonthsFromNow=12;
$dayCondition="Second Friday of";
$months = array();
$years = array();
$currentMonth = (int)date('m');
for($i = $currentMonth; $i < $currentMonth+$noOfMonthsFromNow; $i++) {
$months[] = date('F', mktime(0, 0, 0, $i, 1));
$years[] = date('Y', mktime(0, 0, 0, $i, 1));
}
for ($i=0;$i<count($months);$i++){
$d = date_create($dayCondition.' '.$months[$i].' '.$years[$i]);
if($d instanceof DateTime) echo $d->format('l F d Y H:i:s').'<br>';
}
?>
This can be tested at: http://www.phpfiddle.org/lite/
$beginningOfMonth = strtotime(date('Y-m-01')); // this will give you the timestamp of the beginning of the month
$numTuesdaysPassed = 0;
for ($i = 0; $i <= date('d'); $i ++) { // 'd' == current day of month might need to change to = from <= depending on your needs
if (date('w', $beginningOfMonth + 3600 * $i) == 2) $numTuesdaysPassed ++; // 3600 being seconds in a day, 2 being tuesday from the 'w' (sunday == 0)
}
Not sure if this will work, and there's probably a better way to do it; don't have the means to test it right now but hopefully this puts you on the right track! (I get tripped up on date math a bit too, especially with timezones)
I need to find date x such that it is n working days prior to date y.
I could use something like date("Y-m-d",$def_date." -5 days");, but in that case it wont take into consideration the weekend or off-date. Let's assume my working days would be Monday to Saturday, any idea how I can accomplish this?
Try this
<?php
function businessdays($begin, $end) {
$rbegin = is_string($begin) ? strtotime(strval($begin)) : $begin;
$rend = is_string($end) ? strtotime(strval($end)) : $end;
if ($rbegin < 0 || $rend < 0)
return 0;
$begin = workday($rbegin, TRUE);
$end = workday($rend, FALSE);
if ($end < $begin) {
$end = $begin;
$begin = $end;
}
$difftime = $end - $begin;
$diffdays = floor($difftime / (24 * 60 * 60)) + 1;
if ($diffdays < 7) {
$abegin = getdate($rbegin);
$aend = getdate($rend);
if ($diffdays == 1 && ($astart['wday'] == 0 || $astart['wday'] == 6) && ($aend['wday'] == 0 || $aend['wday'] == 6))
return 0;
$abegin = getdate($begin);
$aend = getdate($end);
$weekends = ($aend['wday'] < $abegin['wday']) ? 1 : 0;
} else
$weekends = floor($diffdays / 7);
return $diffdays - ($weekends * 2);
}
function workday($date, $begindate = TRUE) {
$adate = getdate($date);
$day = 24 * 60 * 60;
if ($adate['wday'] == 0) // Sunday
$date += $begindate ? $day : -($day * 2);
return $date;
}
$def_date="";//define your date here
$preDay='5 days';//no of previous days
date_sub($date, date_interval_create_from_date_string($preDay));
echo businessdays($date, $def_date); //date prior to another date
?>
Modified from PHP.net
Thanks for the help guys, but to solve this particular problem I wrote a simple code:
$sh_padding = 5; //No of working days to count backwards
$temp_sh_padding = 1; //A temporary holder
$end_stamp = strtotime(date("Y-m-d", strtotime($date_format)) . " -1 day"); //The date(timestamp) from which to count backwards
$start_stamp = $end_stamp; //start from same as end day
while($temp_sh_padding<$sh_padding)
{
$sh_day = date('w',$start_stamp);
if($sh_day==0){ //Skip if sunday
}
else
{
$temp_sh_padding++;
}
$start_stamp = strtotime(date("Y-m-d",$start_stamp)." -1 day");
}
$sh_st_dte = date("Y-m-d",$start_stamp); //The required start day
A quick bit of googling got me to this page, which includes a function for calculating the number of working days between two dates.
It should be fairly trivial to adjust that concept to suit your needs.
Your problem, however, is that the concept of "working days" being monday to friday is not universal. If your software is only ever being used in-house, then it's okay to make some assumptions, but if it's intended for use by third parties, then you can't assume that they'll have the same working week as you.
In addition, public holidays will throw a big spanner in the works, by removing arbitrary dates from various working weeks throughout the year.
If you want to cater for these, then the only sensible way of doing it is to store the dates of the year in a calendar (ie a big array), and mark them individually as working or non-working days. And if you're going to do that, then you may as well use the same mechanism for weekends too.
The down-side, of course, is that this would need to be kept up-to-date. But for weekends, at least, that would be trivial (loop through the calendar in advance and mark weekend days where date('w')==0 or date('w')==6).
I want to take a date and work out its week number.
So far, I have the following. It is returning 24 when it should be 42.
<?php
$ddate = "2012-10-18";
$duedt = explode("-",$ddate);
$date = mktime(0, 0, 0, $duedt[2], $duedt[1],$duedt[0]);
$week = (int)date('W', $date);
echo "Weeknummer: ".$week;
?>
Is it wrong and a coincidence that the digits are reversed? Or am I nearly there?
Today, using PHP's DateTime objects is better:
<?php
$ddate = "2012-10-18";
$date = new DateTime($ddate);
$week = $date->format("W");
echo "Weeknummer: $week";
It's because in mktime(), it goes like this:
mktime(hour, minute, second, month, day, year);
Hence, your order is wrong.
<?php
$ddate = "2012-10-18";
$duedt = explode("-", $ddate);
$date = mktime(0, 0, 0, $duedt[1], $duedt[2], $duedt[0]);
$week = (int)date('W', $date);
echo "Weeknummer: " . $week;
?>
$date_string = "2012-10-18";
echo "Weeknummer: " . date("W", strtotime($date_string));
Use PHP's date function
http://php.net/manual/en/function.date.php
date("W", $yourdate)
This get today date then tell the week number for the week
<?php
$date=date("W");
echo $date." Week Number";
?>
Just as a suggestion:
<?php echo date("W", strtotime("2012-10-18")); ?>
Might be a little simpler than all that lot.
Other things you could do:
<?php echo date("Weeknumber: W", strtotime("2012-10-18 01:00:00")); ?>
<?php echo date("Weeknumber: W", strtotime($MY_DATE)); ?>
Becomes more difficult when you need year and week.
Try to find out which week is 01.01.2017.
(It is the 52nd week of 2016, which is from Mon 26.12.2016 - Sun 01.01.2017).
After a longer search I found
strftime('%G-%V',strtotime("2017-01-01"))
Result: 2016-52
https://www.php.net/manual/de/function.strftime.php
ISO-8601:1988 week number of the given year, starting with the first week of the year with at least 4 weekdays, with Monday being the start of the week. (01 through 53)
The equivalent in mysql is DATE_FORMAT(date, '%x-%v')
https://www.w3schools.com/sql/func_mysql_date_format.asp
Week where Monday is the first day of the week (01 to 53).
Could not find a corresponding solution with DateTime.
At least not without solutions like "+1day, last monday".
Edit: since strftime is now deprecated, maybe you can also use date.
Didn't verify it though.
date('o-W',strtotime("2017-01-01"));
I have tried to solve this question for years now, I thought I found a shorter solution but had to come back again to the long story. This function gives back the right ISO week notation:
/**
* calcweek("2018-12-31") => 1901
* This function calculates the production weeknumber according to the start on
* monday and with at least 4 days in the new year. Given that the $date has
* the following format Y-m-d then the outcome is and integer.
*
* #author M.S.B. Bachus
*
* #param date-notation PHP "Y-m-d" showing the data as yyyy-mm-dd
* #return integer
**/
function calcweek($date) {
// 1. Convert input to $year, $month, $day
$dateset = strtotime($date);
$year = date("Y", $dateset);
$month = date("m", $dateset);
$day = date("d", $dateset);
$referenceday = getdate(mktime(0,0,0, $month, $day, $year));
$jan1day = getdate(mktime(0,0,0,1,1,$referenceday[year]));
// 2. check if $year is a leapyear
if ( ($year%4==0 && $year%100!=0) || $year%400==0) {
$leapyear = true;
} else {
$leapyear = false;
}
// 3. check if $year-1 is a leapyear
if ( (($year-1)%4==0 && ($year-1)%100!=0) || ($year-1)%400==0 ) {
$leapyearprev = true;
} else {
$leapyearprev = false;
}
// 4. find the dayofyearnumber for y m d
$mnth = array(0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334);
$dayofyearnumber = $day + $mnth[$month-1];
if ( $leapyear && $month > 2 ) { $dayofyearnumber++; }
// 5. find the jan1weekday for y (monday=1, sunday=7)
$yy = ($year-1)%100;
$c = ($year-1) - $yy;
$g = $yy + intval($yy/4);
$jan1weekday = 1+((((intval($c/100)%4)*5)+$g)%7);
// 6. find the weekday for y m d
$h = $dayofyearnumber + ($jan1weekday-1);
$weekday = 1+(($h-1)%7);
// 7. find if y m d falls in yearnumber y-1, weeknumber 52 or 53
$foundweeknum = false;
if ( $dayofyearnumber <= (8-$jan1weekday) && $jan1weekday > 4 ) {
$yearnumber = $year - 1;
if ( $jan1weekday = 5 || ( $jan1weekday = 6 && $leapyearprev )) {
$weeknumber = 53;
} else {
$weeknumber = 52;
}
$foundweeknum = true;
} else {
$yearnumber = $year;
}
// 8. find if y m d falls in yearnumber y+1, weeknumber 1
if ( $yearnumber == $year && !$foundweeknum) {
if ( $leapyear ) {
$i = 366;
} else {
$i = 365;
}
if ( ($i - $dayofyearnumber) < (4 - $weekday) ) {
$yearnumber = $year + 1;
$weeknumber = 1;
$foundweeknum = true;
}
}
// 9. find if y m d falls in yearnumber y, weeknumber 1 through 53
if ( $yearnumber == $year && !$foundweeknum ) {
$j = $dayofyearnumber + (7 - $weekday) + ($jan1weekday - 1);
$weeknumber = intval( $j/7 );
if ( $jan1weekday > 4 ) { $weeknumber--; }
}
// 10. output iso week number (YYWW)
return ($yearnumber-2000)*100+$weeknumber;
}
I found out that my short solution missed the 2018-12-31 as it gave back 1801 instead of 1901. So I had to put in this long version which is correct.
How about using the IntlGregorianCalendar class?
Requirements: Before you start to use IntlGregorianCalendar make sure that libicu or pecl/intl is installed on the Server.
So run on the CLI:
php -m
If you see intl in the [PHP Modules] list, then you can use IntlGregorianCalendar.
DateTime vs IntlGregorianCalendar:
IntlGregorianCalendar is not better then DateTime. But the good thing about IntlGregorianCalendar is that it will give you the week number as an int.
Example:
$dateTime = new DateTime('21-09-2020 09:00:00');
echo $dateTime->format("W"); // string '39'
$intlCalendar = IntlCalendar::fromDateTime ('21-09-2020 09:00:00');
echo $intlCalendar->get(IntlCalendar::FIELD_WEEK_OF_YEAR); // integer 39
<?php
$ddate = "2012-10-18";
$duedt = explode("-",$ddate);
$date = mktime(0, 0, 0, $duedt[1], $duedt[2],$duedt[0]);
$week = (int)date('W', $date);
echo "Weeknummer: ".$week;
?>
You had the params to mktime wrong - needs to be Month/Day/Year, not Day/Month/Year
To get the week number for a date in North America I do like this:
function week_number($n)
{
$w = date('w', $n);
return 1 + date('z', $n + (6 - $w) * 24 * 3600) / 7;
}
$n = strtotime('2022-12-27');
printf("%s: %d\n", date('D Y-m-d', $n), week_number($n));
and get:
Tue 2022-12-27: 53
for get week number in jalai calendar you can use this:
$weeknumber = date("W"); //number week in year
$dayweek = date("w"); //number day in week
if ($dayweek == "6")
{
$weeknumberint = (int)$weeknumber;
$date2int++;
$weeknumber = (string)$date2int;
}
echo $date2;
result:
15
week number change in saturday
The most of the above given examples create a problem when a year has 53 weeks (like 2020). So every fourth year you will experience a week difference. This code does not:
$thisYear = "2020";
$thisDate = "2020-04-24"; //or any other custom date
$weeknr = date("W", strtotime($thisDate)); //when you want the weeknumber of a specific week, or just enter the weeknumber yourself
$tempDatum = new DateTime();
$tempDatum->setISODate($thisYear, $weeknr);
$tempDatum_start = $tempDatum->format('Y-m-d');
$tempDatum->setISODate($thisYear, $weeknr, 7);
$tempDatum_end = $tempDatum->format('Y-m-d');
echo $tempDatum_start //will output the date of monday
echo $tempDatum_end // will output the date of sunday
Very simple
Just one line:
<?php $date=date("W"); echo "Week " . $date; ?>"
You can also, for example like I needed for a graph, subtract to get the previous week like:
<?php $date=date("W"); echo "Week " . ($date - 1); ?>
Your code will work but you need to flip the 4th and the 5th argument.
I would do it this way
$date_string = "2012-10-18";
$date_int = strtotime($date_string);
$date_date = date($date_int);
$week_number = date('W', $date_date);
echo "Weeknumber: {$week_number}.";
Also, your variable names will be confusing to you after a week of not looking at that code, you should consider reading http://net.tutsplus.com/tutorials/php/why-youre-a-bad-php-programmer/
The rule is that the first week of a year is the week that contains the first Thursday of the year.
I personally use Zend_Date for this kind of calculation and to get the week for today is this simple. They have a lot of other useful functions if you work with dates.
$now = Zend_Date::now();
$week = $now->get(Zend_Date::WEEK);
// 10
To get Correct Week Count for Date 2018-12-31 Please use below Code
$day_count = date('N',strtotime('2018-12-31'));
$week_count = date('W',strtotime('2018-12-31'));
if($week_count=='01' && date('m',strtotime('2018-12-31'))==12){
$yr_count = date('y',strtotime('2018-12-31')) + 1;
}else{
$yr_count = date('y',strtotime('2018-12-31'));
}
function last_monday($date)
{
if (!is_numeric($date))
$date = strtotime($date);
if (date('w', $date) == 1)
return $date;
else
return date('Y-m-d',strtotime('last monday',$date));
}
$date = '2021-01-04'; //Enter custom date
$year = date('Y',strtotime($date));
$date1 = new DateTime($date);
$ldate = last_monday($year."-01-01");
$date2 = new DateTime($ldate);
$diff = $date2->diff($date1)->format("%a");
$diff = $diff/7;
$week = intval($diff) + 1;
echo $week;
//Returns 2.
try this solution
date( 'W', strtotime( "2017-01-01 + 1 day" ) );