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.
So I have a script that returns the number of weeks in a particular month and year. How can I take a specific day from that month and determine if it is part of week 1,2,3,4 or 5 of that month?
The most frustrating thing I have ever tried to get working - but here it is!
<?php
/**
* Returns the amount of weeks into the month a date is
* #param $date a YYYY-MM-DD formatted date
* #param $rollover The day on which the week rolls over
*/
function getWeeks($date, $rollover)
{
$cut = substr($date, 0, 8);
$daylen = 86400;
$timestamp = strtotime($date);
$first = strtotime($cut . "00");
$elapsed = ($timestamp - $first) / $daylen;
$weeks = 1;
for ($i = 1; $i <= $elapsed; $i++)
{
$dayfind = $cut . (strlen($i) < 2 ? '0' . $i : $i);
$daytimestamp = strtotime($dayfind);
$day = strtolower(date("l", $daytimestamp));
if($day == strtolower($rollover)) $weeks ++;
}
return $weeks;
}
//
echo getWeeks("2011-06-11", "sunday"); //outputs 2, for the second week of the month
?>
Edit: so much for "single line" - needed variables to avoid recomputation with the conditional. Tossed in a default argument while I was at it.
function weekOfMonth($when = null) {
if ($when === null) $when = time();
$week = date('W', $when); // note that ISO weeks start on Monday
$firstWeekOfMonth = date('W', strtotime(date('Y-m-01', $when)));
return 1 + ($week < $firstWeekOfMonth ? $week : $week - $firstWeekOfMonth);
}
Please note that weekOfMonth(strtotime('Oct 31, 2011')); will return 6; some rare months have 6 weeks in them, contrary to OP's expectation. January 2017 is another month with 6 ISO weeks - Sunday the 1st falls in the last year's week, since ISO weeks start on Monday.
For starshine531, to return a 0 indexed week of the month, change the return 1 + to return 0 + or return (int).
For Justin Stayton, for weeks starting on Sunday instead of Monday I would use strftime('%U' instead of date('W', as follows:
function weekOfMonth($when = null) {
if ($when === null) $when = time();
$week = strftime('%U', $when); // weeks start on Sunday
$firstWeekOfMonth = strftime('%U', strtotime(date('Y-m-01', $when)));
return 1 + ($week < $firstWeekOfMonth ? $week : $week - $firstWeekOfMonth);
}
For this version, 2017-04-30 is now in week 6 of April, while 2017-01-31 is now in week 5.
public function getWeeks($timestamp)
{
$maxday = date("t",$timestamp);
$thismonth = getdate($timestamp);
$timeStamp = mktime(0,0,0,$thismonth['mon'],1,$thismonth['year']); //Create time stamp of the first day from the give date.
$startday = date('w',$timeStamp); //get first day of the given month
$day = $thismonth['mday'];
$weeks = 0;
$week_num = 0;
for ($i=0; $i<($maxday+$startday); $i++) {
if(($i % 7) == 0){
$weeks++;
}
if($day == ($i - $startday + 1)){
$week_num = $weeks;
}
}
return $week_num;
}
Hello all i have been struggling for the whole day trying to figure this code out, i finally figured it out so i thought i would share it with you all.
all you need to do is put a time stamp into the function and it will return the week number back to you.
thanks
there is a problem with this method. if the passing date (Lets say 2012/01/01 which is a Sunday) and "$rollover" day is "Sunday", then this function will return 2. where its actually is 1'st week. i think i have fixed it in following function.
please add comments to make it better.
function getWeeks($date, $rollover)
{
$cut = substr($date, 0, 8);
$daylen = 86400;
$timestamp = strtotime($date);
$first = strtotime($cut . "01");
$elapsed = (($timestamp - $first) / $daylen)+1;
$i = 1;
$weeks = 0;
for($i==1; $i<=$elapsed; $i++)
{
$dayfind = $cut . (strlen($i) < 2 ? '0' . $i : $i);
$daytimestamp = strtotime($dayfind);
$day = strtolower(date("l", $daytimestamp));
if($day == strtolower($rollover))
{
$weeks++;
}
}
if($weeks==0)
{
$weeks++;
}
return $weeks;
}
This is a solution based on sberry's mathematical solution but using the PHP DateTime class instead.
function week_of_month($date) {
$first_of_month = new DateObject($date->format('Y/m/1'));
$day_of_first = $first_of_month->format('N');
$day_of_month = $date->format('j');
return floor(($day_of_first + $day_of_month - 1) / 7) + 1;
}
Just Copy and Past the code and pass month and year.
e.g month=04 year=2013.
That's exactly what You Need.
$mm= $_REQUEST['month'];
$yy= $_REQUEST['year'];
$startdate=date($yy."-".$mm."-01") ;
$current_date=date('Y-m-t');
$ld= cal_days_in_month(CAL_GREGORIAN, $mm, $yy);
$lastday=$yy.'-'.$mm.'-'.$ld;
$start_date = date('Y-m-d', strtotime($startdate));
$end_date = date('Y-m-d', strtotime($lastday));
$end_date1 = date('Y-m-d', strtotime($lastday." + 6 days"));
$count_week=0;
$week_array = array();
for($date = $start_date; $date <= $end_date1; $date = date('Y-m-d', strtotime($date. ' + 7 days')))
{
$getarray=getWeekDates($date, $start_date, $end_date);
echo "<br>";
$week_array[]=$getarray;
echo "\n";
$count_week++;
}
// its give the number of week for the given month and year
echo $count_week;
//print_r($week_array);
function getWeekDates($date, $start_date, $end_date)
{
$week = date('W', strtotime($date));
$year = date('Y', strtotime($date));
$from = date("Y-m-d", strtotime("{$year}-W{$week}+1"));
if($from < $start_date) $from = $start_date;
$to = date("Y-m-d", strtotime("{$year}-W{$week}-6"));
if($to > $end_date) $to = $end_date;
$array1 = array(
"ssdate" => $from,
"eedate" => $to,
);
return $array1;
// echo "Start Date-->".$from."End Date -->".$to;
}
for($i=0;$i<$count_week;$i++)
{
$start= $week_array[$i]['ssdate'];
echo "--";
$week_array[$i]['eedate'];
echo "<br>";
}
OUTPUT:
week( 0 )=>2013-03-01---2013-03-02
week( 1 )=>2013-03-03---2013-03-09
week( 2 )=>2013-03-10---2013-03-16
week( 3 )=>2013-03-17---2013-03-23
week( 4 )=>2013-03-24---2013-03-30
week( 5 )=>2013-03-31---2013-03-31
I think I found an elegant solution
$time = time(); // or whenever
$week_of_the_month = ceil(date('d', $time)/7);
For a Monday-Sunday (ISO 8601) week (or, if you simply don't care), you can do this in one line:
function get_week_of_month($date) {
return date('W', $date) - date('W', strtotime(date("Y-m-01", $date))) + 1;
}
(Source)
For anything else, (e.g. a Sunday-Saturday week), you just need to tweak $date inside the function:
function get_week_of_month($date) {
$date += 86400; //For weeks starting on Sunday
return date('W', $date) - date('W', strtotime(date("Y-m-01", $date))) + 1;
}
(Thanks to these guys/gals)
NOTE: You may run into some issues at the end of the year (e.g. around 12/31, 1/1, etc.). Read more here.
This is the snippet that I made to fulfill my requirements for the same. Hope this will help you.
function getWeek($timestamp) {
$week_year = date('W',$timestamp);
$week = 0;//date('d',$timestamp)/7;
$year = date('Y',$timestamp);
$month = date('m',$timestamp);
$day = date('d',$timestamp);
$prev_month = date('m',$timestamp) -1;
if($month != 1 ){
$last_day_prev = $year."-".$prev_month."-1";
$last_day_prev = date('t',strtotime($last_day_prev));
$week_year_last_mon = date('W',strtotime($year."-".$prev_month."-".$last_day_prev));
$week_year_first_this = date('W',strtotime($year."-".$month."-1"));
if($week_year_first_this == $week_year_last_mon){
$week_diff = 0;
}
else{
$week_diff = 1;
}
if($week_year ==1 && $month == 12 ){
// to handle December's last two days coming in first week of January
$week_year = 53;
}
$week = $week_year-$week_year_last_mon + 1 +$week_diff;
}
else{
// to handle first three days January coming in last week of December.
$week_year_first_this = date('W',strtotime($year."-01-1"));
if($week_year_first_this ==52 || $week_year_first_this ==53){
if($week_year == 52 || $week_year == 53){
$week =1;
}
else{
$week = $week_year + 1;
}
}
else{
$week = $week_year;
}
}
return $week;
}
This is probably not a good way to do this but it's my first thought and I'm really tired.
Put all your dates into an array. The date object must have a day name (Monday). Create a method that searches the array and when ever you hit a Sunday you add 1 to a week counter. Once you find the date you're looking for return the week counter. That is the week the day falls in of the year. For the week in the month you have to reset the week counter every time you get to the last day in each month.
Here comes two liner:
function getWeekOfMonth(DateTime $date) {
$firstDayOfMonth = new DateTime($date->format('Y-m-1'));
return ceil(($firstDayOfMonth->format('N') + $date->format('j') - 1) / 7);
}
And Wtower's solutions doesn't work 100% properly.
Thought I'd share my function as well. This returns an array of weeks. Every week is an array with weeks day (0..6) as key and months day (1..31) as value.
Function assumes that week starts with Sunday.
Enjoy!
function get_weeks($year, $month){
$days_in_month = date("t", mktime(0, 0, 0, $month, 1, $year));
$weeks_in_month = 1;
$weeks = array();
//loop through month
for ($day=1; $day<=$days_in_month; $day++) {
$week_day = date("w", mktime(0, 0, 0, $month, $day, $year));//0..6 starting sunday
$weeks[$weeks_in_month][$week_day] = $day;
if ($week_day == 6) {
$weeks_in_month++;
}
}
return $weeks;
}
My 5 cents:
/**
* calculate number of weeks in a particular month
*/
function weeksInMonth($month=null,$year=null){
if( null==($year) ) {
$year = date("Y",time());
}
if(null==($month)) {
$month = date("m",time());
}
// find number of days in this month
$daysInMonths = date('t',strtotime($year.'-'.$month.'-01'));
$numOfweeks = ($daysInMonths%7==0?0:1) + intval($daysInMonths/7);
$monthEndingDay= date('N',strtotime($year.'-'.$month.'-'.$daysInMonths));
$monthStartDay = date('N',strtotime($year.'-'.$month.'-01'));
if($monthEndingDay<$monthStartDay){
$numOfweeks++;
}
return $numOfweeks;
}
I create this function, from brazil :) I hope it is useful
function weekofmonth($time) {
$firstday = 1;
$lastday = date('j',$time);
$lastdayweek = 6; //Saturday
$week = 1;
for ($day=1;$day<=$lastday;$day++) {
$timetmp = mktime(0, 0, 0, date('n',$time), $day, date('Y',$time));
if (date('N',$timetmp) == $lastdayweek) {
$week++;
}
}
if (date('N',$time)==$lastdayweek) {
$week--;
}
return $week;
}
$time = mktime(0, 0, 0, 9, 30, 2014);
echo weekofmonth($time);
I found a easy way to determine what week of the month today is in, and it would be a small change to have it work on any other date. I'm adding my two cents in here as I think my way is much more compact then the methods listed.
$monthstart = date("N",strtotime(date("n/1/Y")));
$date =( date("j")+$monthstart ) /7;
$ddate= floor( $date );
if($ddate != date) {$ddate++;}
and $ddate contains the week number you could modify it like so
function findweek($indate)
{
$monthstart = date("N",strtotime(date("n/1/Y",strtotime($indate))));
$date =( date("j",strtotime($indate))+$monthstart ) /7;
$ddate= floor( $date );
if($ddate != $date) {$ddate++;}
return $ddate;
}
and it would return what week of the month any date you give it is.
what it does is first find the number of days from the start of the week to the first of the month. then adds that on to the current date then divides the new date by 7 and that will give you how many weeks have passed since the start of the month, including a decimal place for the part of the the current week that has passed. so what I do next is round down that number, then compare the rounded down version to the original if the two match your at the end of the week so it's already in the number. if they don't then just add one to the rounded down number and voila you have the current week number.
Srahul07's solution works perfectly... If you abide by the Monday-Sunday week system! Here in 'murica, non-business folk tend to go by Sunday-Saturday being a week, so May 1, 2011 is week 1 and May 2, 2011 is still week 1.
Adding the following logic to the bottom of his function, right before it returns $week will convert this to a Sunday -> Monday system:
if (!date('w',strtotime("$year-$month-01")) && date('w',$timestamp))
$week--;
elseif (date('w',strtotime("$year-$month-01")) && !date('w',$timestamp))
$week++;
After alot of efoort i found the solution
<?php
function getWeeks($month,$year)
{
$month = intval($month); //force month to single integer if '0x'
$suff = array('st','nd','rd','th','th','th'); //week suffixes
$end = date('t',mktime(0,0,0,$month,1,$year)); //last date day of month: 28 - 31
$start = date('w',mktime(0,0,0,$month,1,$year)); //1st day of month: 0 - 6 (Sun - Sat)
$last = 7 - $start; //get last day date (Sat) of first week
$noweeks = ceil((($end - ($last + 1))/7) + 1); //total no. weeks in month
$output = ""; //initialize string
$monthlabel = str_pad($month, 2, '0', STR_PAD_LEFT);
for($x=1;$x<$noweeks+1;$x++)
{
if($x == 1)
{
$startdate = "$year-$monthlabel-01";
$day = $last - 6;
}
else
{
$day = $last + 1 + (($x-2)*7);
$day = str_pad($day, 2, '0', STR_PAD_LEFT);
$startdate = "$year-$monthlabel-$day";
}
if($x == $noweeks)
{
$enddate = "$year-$monthlabel-$end";
}
else
{
$dayend = $day + 6;
$dayend = str_pad($dayend, 2, '0', STR_PAD_LEFT);
$enddate = "$year-$monthlabel-$dayend";
}
$j=1;
if($j--)
{
$k=getTotalDate($startdate,$enddate);
$j=1;
}
$output .= "Week ".$xyz." week -> Start date=$startdate End date=$enddate <br />";
}
return $output;
}
if(isset($_POST) && !empty($_POST)){
$month = $_POST['m'];
$year = $_POST['y'];
echo getWeeks($month,$year);
}
?>
<form method="post">
M:
<input name="m" value="" />
Y:
<input name="y" value="" />
<input type="submit" value="go" />
</form>
I really liked #michaelc's answer. However, I got stuck on a few points. It seemed that every time Sunday rolled around, there was an offset of one. I think it has to do with what day of the week is the start of the week. In any case, here is my slight alteration to it, expanded a bit for readability:
function wom(\DateTime $date) {
// The week of the year of the current month
$cw = date('W', $date->getTimestamp());
// The week of the year of the first of the given month
$fw = date('W',strtotime(date('Y-m-01',$date->getTimeStamp())));
// Offset
$o = 1;
// If it is a Saturday, offset by two.
if( date('N',$date->getTimestamp()) == 7 ) {
$o = 2;
}
return $cw -$fw + $o;
}
So if the date is Nov. 9, 2013...
$cw = 45
$fw = 44
and with the offset of 1, it correctly returns 2.
If the date is Nov. 10, 2013, $cw and $fw are the same as before, but the offset is 2, and it correctly returns 3.
function get_week_of_month( $timestamp )
{
$week_of_month = 0;
$month = date( 'j', $timestamp );
$test_month = $month;
while( $test_month == $month )
{
$week_of_month++;
$timestamp = strtotime( '-1 week', $timestamp );
$test_month = date( 'j', $timestamp );
}
return $week_of_month;
}
I found this online:
http://kcwebprogrammers.blogspot.de/2009/03/current-week-in-month-php.html
He has a very simple solution which seems to work fine for me.
$currentWeek = ceiling((date("d") - date("w") - 1) / 7) + 1;
So for example:
$now = strtotime("today");
$weekOfMonth = ceil((date("d", $now) - date("w", $now) - 1) / 7) + 1;
you can use W in newer php versions. http://php.net/manual/en/function.date.php
i have used it like so:
function getWeek($date) {
$month_start=strtotime("1 ".date('F Y',$date));
$current_date=strtotime(date('j F Y',$date));
$month_week=date("W",$month_start);
$current_week=date("W",$current_date);
return ($current_week-$month_week);
}//0 is the week of the first.
Short and foolproof:
// Function accepts $date as a string,
// Returns the week number in which the given date falls.
// Assumed week starts on Sunday.
function wom($date) {
$date = strtotime($date);
$weeknoofday = date('w', $date);
$day = date('j', $date);
$weekofmonth = ceil(($day + (7-($weeknoofday+1))) / 7);
return $weekofmonth;
}
// Test
foreach (range(1, 31) as $day) {
$test_date = "2015-01-" . str_pad($day, 2, '0', STR_PAD_LEFT);
echo "$test_date - ";
echo wom($test_date) . "\n";
}
I use this simple function:
function weekNumberInMonth($timestampDate)
{
$firstDayOfMonth = strtotime(date('01-M-Y 00:00:00', $timestampDate));
$firstWeekdayOfMonth = date( 'w', $firstDayOfMonth);
$dayNumberInMonth = date('d', $timestampDate);
$weekNumberInMonth = ceil(($dayNumberInMonth + $firstWeekdayOfMonth) / 7);
return $weekNumberInMonth;
}
if I understand correct, the question is how to identify what number of week within a month of a specific day... I was looking for similar solution. I used some ideas of above answers to develop my own solution. Hope it can be helpful for somebody. If Yes, then UpVote my answer.
function week_number_within_month($datenew){
$year = date("Y",strtotime($datenew));
$month = date("m",strtotime($datenew));
// find number of days in this month
$daysInMonths = date('t',strtotime($year.'-'.$month.'-01'));
$numOfweeks = ($daysInMonths%7==0?0:1) + intval($daysInMonths/7);
$monthEndingDay= date('N',strtotime($year.'-'.$month.'-'.$daysInMonths));
$monthStartDay = date('N',strtotime($year.'-'.$month.'-01'));
if($monthEndingDay<$monthStartDay){
$numOfweeks++;
}
$date=date('Y/m/d', strtotime($year.'-'. $month.'-01'));
$week_array=Array();
for ($i=1; $i<=$numOfweeks; $i++){ /// create an Array of all days of month separated by weeks as a keys
$max = 7;
if ($i ==1){ $max = 8 - $monthStartDay;}
if ($i == $numOfweeks){ $max = $monthEndingDay;}
for ($r=1; $r<=$max; $r++){
$week_array[$i][]=$date;
$date = date('Y/m/d',strtotime($date . "+1 days"));
}
}
$new_datenew = date('Y/m/d', strtotime($datenew));
$week_result='';
foreach ($week_array as $key => $val){ /// finding what week number of my date from week_array
foreach ($val as $kr => $value){
if ($new_datenew == $value){
$week_result = $key;
}
}
}
return $week_result;
}
print week_number_within_month('2016-09-15');
function getWeekOfMonth(\DateTime $date)
{
$firstWeekdayOfMonth = new DateTime("first weekday 0 {$date->format('M')} {$date->format('Y')}");
$offset = $firstWeekdayOfMonth->format('N')-1;
return intval(($date->format('j') + $offset)/7)+1;
}
/**
* In case of Week we can get the week of year. So whenever we will get the week of the month then we have to
* subtract the until last month weeks from it will give us the current month week.
*/
$dateComponents = getdate();
if($dateComponents['mon'] == 1)
$weekOfMonth = date('W', strtotime($dateComponents['year'].'-'.$dateComponents['mon'].'-'.$dateComponents['mday']))-1; // We subtract -1 to map it to the array
else
$weekOfMonth = date('W', strtotime($dateComponents['year'].'-'.$dateComponents['mon'].'-'.$dateComponents['mday']))-date('W', strtotime($dateComponents['year'].'-'.$dateComponents['mon'].'-01'));
Using Carbon:
$date = Carbon::now();
$d1 = $date->startOfMonth();
$d2 = $date->endOfMonth();
$weeks = $d1->diffInWeeks($d2);
If you clearly want to separate a month into 4 Weeks, you can use this function.
This is helpful, if you want
"the first monday of month"
"the third thursday of month" etc.
Here we go
/**
* This Calculates (and returns) the week number within a month, based on date('j') day of month.
* This is useful, if you want to have (for instance) the first Thu in month, regardless of date
* #param $Timestamp
* #return float|int
*/
function getWeekOfMonth($Timestamp)
{
$DayOfMonth=date('j', $Timestamp); // Day of the month without leading zeros 0-31
if($DayOfMonth>21) return 4;
if($DayOfMonth>14) return 3;
if($DayOfMonth>7) return 2;
return 1;
}
From carbon:
return (int) ceil((new Datetime())->format('d') / 7);
As simple as possible :)
Python: Number of the Week in a Month
This is a worked example in Python - should be simple to convert.
for($x=0; $x<12; $x++)
{
$month = mktime(0, 0, 0, date("m")+$x, date("d"), date("Y"));
$key = date('m', $month);
$monthname = date('F', $month);
$months[$key] = $monthname;
}
I know for sure I'm doing the math incorrectly for the 4th parameter of mktime. I'm starting with the current month number ( 7 being July ) and adding 1 for each next month, sometimes it ends up being that the same month is returned twice, maybe because I'm not setting it to the beginning of the month? How would you improve/recode this?
Result is that $months would result in an array where 07 = July 08 = August, 09 = September. Right now it populates October twice. I think it has to do with today being the 31st and it incorrectly adds and reaches the next month.
Just fixed your code slightly, this should work pretty well:
$months = array();
$currentMonth = (int)date('m');
for ($x = $currentMonth; $x < $currentMonth + 12; $x++) {
$months[] = date('F', mktime(0, 0, 0, $x, 1));
}
Note that I took out the array key, as I think it's unnecessary, but you can change that of course if you need it.
An alternative would be to use strtotime:
for ($x=0; $x < 12; $x++) {
$time = strtotime('+' . $x . ' months', strtotime(date('Y-M' . '-01')));
$key = date('m', $time);
$name = date('F', $time);
$months[$key] = $name;
}
In my opinion this code is easier to read.
Less complicated, no loops, generic array keys:
function stackoverflow_get_monthname($x){
return date("F",mktime(NULL, NULL, NULL, (int)date("n") + ($x+1), NULL, NULL));
}
$months = array_map("stackoverflow_get_monthname", range(1,12) );
var_dump($months);
Given 2592000 is 30 days.
$month_time = 60*60*24*30; // 30 Days
for($x=0; x<12; $x++)
{
$time = time()+($month_time*$x);
$key = date('m', $time);
$month[$key] = date('F', $time);
}
In an answer on StackOverflow, can't find it right now, someone compared the performance of multiple methods of creating a time 1 week from now. Directly using numbers was much more efficient than any other method.
You might be getting the last day of the month (the 31st) bug - which led to two months with the same link - that Eddy very nicely figured out for me with this answer:
$current_month = date('n');
$MONTHS = array();
for ($m=0; $m<12; $m++) {
$display_month = $m + $current_month;
$MONTHS[] = date('F',mktime(1,1,1,$display_month,1,date("Y")));
"Result is that $months would result in an array where 07 = July 08 = August, 09 = September."
for ($key = 1; $key <=12; $key++) {
$months[str_pad($key, 2, '0', STR_PAD_LEFT)] = date('F', strtotime('2000-' . $key));
}
If you're okay with 7 = July 8 = August, 9 = September, then:
for ($key = 1; $key <=12; $key++) {
$months[$key] = date('F', strtotime('2000-' . $key));
}
Sometime you need to be careful on your locale, so this is my solution (in a function):
$months = [];
for ($x=1; $x < 13; $x++) {
$time = mktime(0, 0, 0, $x, 1);
$key = date('m', $time);
$name = ucfirst(strftime('%B', $time));
$months[(int)$key] = $name;
}
return $months;
Here is a simple script to go forward 12 months from the current date. It includes the year with it.
# Set Number of Months to Traverse
$num_months = 12;
# Set Current Month as the 1st
$current_month = date('Y-m').'-01';
for ($count = 0; $count <= $num_months; $count++) {
# Fetch Date for each as YYYY-MM-01
$dates[] = date('Y-m', strtotime($current_month.' + '.$count.' Months')).'-01';
}
You could turn this into a select list with the current month selected by dropping this in:
echo '<select name="month">';
foreach ($dates as $d) {
echo '<option value="'.$d.'"';
if ($d == date('Y-m').'-01') echo 'selected="selected"';
echo '>'.date('F, Y', strtotime($d)).'</option>';
}
echo '</select>';