Consider the following code that was working perfectly yesterday, but now today (30th March), it has stopped working...
<form class="date">
<select name="date" onchange="this.form.submit()">
<?php
for ($i = 0; $i <= 60; ++$i) {
$time = strtotime(sprintf('+%d months', $i));
$value = date('Y-m', $time);
$label = date('F Y', $time);
$nowvalue = date('Y-m');
$nowlabel = date('F Y');
if(isset($_GET['date'])) {
if(strcmp($value,$_GET['date']) == 0) {
//If 0, $value and $_GET['date'] are the same: The option is selected
printf('<option value="%s" selected>%s</option>', $value, $label);
} else {
printf('<option value="%s">%s</option>', $value, $label);
}
} else {
if($value == $nowvalue) {
printf('<option value="%s" selected>%s</option>', $value, $label);
} else {
printf('<option value="%s">%s</option>', $value, $label);
}
}
}
?>
</select>
</form>
This outputs a select list of upcoming months, that onchange, reloads the page with a different query in the URL for the calendar on that page to react on.
If i manually update the query in the URL, it works fine, but this select list is repeating dates, have a look at the image below...
This cant be a coincidence that all of the months with the shortest days are missing, or in fact have been replaced with the same name as the month before?
Does anyone have an idea why this might be happening?
This is because when you loop through dates on a monthly basis on the 31st of the month you will come across months that have less than 30 days. PHP does automatically adjust the date to the last day of the month but instead goes to the end of the month and then adds the remaining days to it and gets the date that way which is the next month.
So if you get to February, which only has 28 days (most years) it will get to the 28th and add three days and give you March 3rd.
To solve this you need to set the date to the first day of the month and then iterate through the dates.
This is just an FYI as I realise that you've added your own answer.
You could instead use DateTime() to simplify your code (if you want).
So you could end up with something like:
<select name="date" onchange="this.form.submit()">
<?php
$selected = DateTime::createFromFormat('Y-m-d', #$_GET['date'] . '-01') ?: new DateTime('first day of this month');
for ($i = 0; $i <= 60; $i++) {
$date = (new DateTime('first day of this month'))->modify("+$i months");
$select = $date == $selected ? 'selected' : '';
printf('<option value="%s" %s>%s</option>', $date->format('Y-m'), $select, $date->format('F-Y'));
}
?>
</select>
Hope this helps!
As John Conde pointed out, this is because todays dates (the 31st) plus one month is actual May 1st, and April is skipped.
I did the following to set the date to the 1st of the month, then plus one month...
<?php
for ($i = 0; $i <= 60; ++$i) {
$firstofthismonth = '01-' . date('m-Y');
$time = strtotime(sprintf($firstofthismonth . '+%d months', $i));
$value = date('Y-m', $time);
$label = date('F Y', $time);
$nowvalue = date('Y-m');
$nowlabel = date('F Y');
if(isset($_GET['date'])) {
if(strcmp($value,$_GET['date']) == 0) {
//If 0, $value and $_GET['date'] are the same: The option is selected
printf('<option value="%s" selected>%s</option>', $value, $label);
} else {
printf('<option value="%s">%s</option>', $value, $label);
}
} else {
if($value == $nowvalue) {
printf('<option value="%s" selected>%s</option>', $value, $label);
} else {
printf('<option value="%s">%s</option>', $value, $label);
}
}
}
?>
does anyone know how I would echo out the selected date as text with the date month and year separated outside of the form? I tried echoing out $date $month and $year outside of the form however this doesn't give me the correct date thankyou for the help
<?
$date = array('16-01-14','16-01-28','16-02-14','16-02-28','16-03-14','16-03-28','16-04-14','16-04-28',
'16-05-14','16-05-28','16-06-14','16-06-28','16-07-14','16-07-28','16-08-14','16-08-28','16-09-14','16-09-28','16-10-14','16-10-28',
'16-11-14','16-11-28','16-12-14','16-12-28');
$currentdate = date('y-m-d');
echo $currentdate;
?>
<form>
<select style="width:200px;">
<?php
foreach ($date as $i => $d) {
if ($currentdate >= $d && ($i == count($date)-1 || $currentdate < $date[$i+1])) {
$selected = "selected";
} else {
$selected = "";
}
list($year, $month, $day) = explode('-', $d);
echo "<option $selected>" . date("m/d/Y", strtotime($d)) . "</option>";
echo 'the current billing period is';
}
?>
</select>
</form>
Inside of your loop add a $selected_int variable like so:
foreach ($date as $i => $d) {
if ($currentdate >= $d && ($i == count($date)-1 || $currentdate < $date[$i+1])) {
$selected = "selected";
$selected_int = $i;
} else {
$selected = "";
}
list($year, $month, $day) = explode('-', $d);
echo "<option $selected>" . date("m/d/Y", strtotime($d)) . "</option>";
echo 'the current billing period is';
}
Then, you can reference it like:
echo date('Y-m-d', strtotime($date[$selected_int]));
Addition
I know you've already accepted the answer, but I also wanted to make a suggestion now that I see what you are using the $date for. Since you know the start date, and it is in 14-day periods, it would be easy to write that as part of the loop.
$start_date = date('Y-m-d', strtotime(date('Y').'-01-01'); //First day of the year, for the sake of argument.
$interval = 14;
for ($i = 0; date('Y') == date('Y', strtotime($start_date.' +'.($i * $interval).' days')); $i++) {//While this year is equal to the start date's year with the added interval [If I knew what your logic here was I could offer a better suggestion]
if ($currentdate >= date("Y-m-d", strtotime($start_date.' +'.($i * $interval).' days')) && (date('Y') < date("Y", strtotime($start_date.' +'.(($i + 1) * $interval).' days')) || $currentdate < date("m/d/Y", strtotime($start_date.' +'.(($i + 1) * $interval).' days')))) {
$selected = "selected";
$selected_int = $i;
} else {
$selected = "";
}
echo "<option $selected>" . date("m/d/Y", strtotime($start_date.' +'.($i * $interval).' days')) . "</option>";
}
Basically, this takes the start date, shows it as the first date option, then adds 14 days to it with each pass through. Your if/else statement should still be the same. It checks to see if you are on the last interval of the year, or if the current date is less than the next interval, and also that the current date is greater than the current interval.
After your loop, you can get the date by:
echo date("m/d/Y", strtotime($start_date.' +'.($selected_int * $interval).' days'));
I know it seems like a lot, but it would save you from having to make a date array to begin with.
Use strtotime instead list.
....
// list($year, $month, $day) = explode('-', $d);
echo "<option $selected>" . date("m/d/Y", strtotime($d)) . "</option>";
....
EDIT: Additional information - your code requires a lot modification and likely some structure changes but assuming this is for testing a method and "how to do" instead a final product.
You need to submit the selected date, catch it in the script and use the selected date to do what you need - i.e. retrieve data from database - and this should give you some idea.
<?php
// You need to create these dates by using another method. You cannot hard code these. You can create it with date functions easily.
$date = array('16-01-14','16-01-28','16-02-14','16-02-28','16-03-14','16-03-28','16-04-14','16-04-28','16-05-14','16-05-28','16-06-14','16-06-28','16-07-14','16-07-28','16-08-14','16-08-28','16-09-14','16-09-28','16-10-14','16-10-28','16-11-14','16-11-28','16-12-14','16-12-28');
// Checking if we have a posted form, with the button name user clicked
if (isset($_POST["btnSubmit"])) {
// This is your selected day - use it where you need:
$selectedDate = $_POST["selectedDate"];
// This is where your model start singing and gets necessary info for this date - just printing here as sample
print $selectedDate;
// I need dropDownDate to compare in the SELECT to preselect the appropriate date
$dropDownDate = strtotime($selectedDate);
} else {
// First time visit, preselect the nearest date by using current date
$dropDownDate = time();
}
?>
<form method="post">
<select name="selectedDate" style="width:200px;">
<?php
foreach ($date as $i => $d) {
if ($dropDownDate >= strtotime($d) &&
(!isset($date[$i+1]) || ($dropDownDate < strtotime($date[$i+1])))
) {
$selected = 'selected="selected"';
} else {
$selected = "";
}
list($year, $month, $day) = explode('-', $d);
echo "<option $selected>" . date("m/d/Y", strtotime($d)) . "</option>";
}
?>
</select>
<input type="submit" name="btnSubmit" value="Submit">
</form>
Note that I added a "submit" type input (to submit the form) and changed form method to "post", finally named SELECT as "selectedDate". I also changed your date comparison code line in the loop.
Hope this helps.
I need to generate an html code of a list of 10 open days following the current day, by open days I mean business days (m,t,w,t and f), I'm using the following function to translate the date to french:
function f_date() {
$temps = time();
$jours = array('Dimanche', 'Lundi', 'Mardi', 'Mercredi', 'Jeudi', 'Vendredi', 'Samedi');
$jours_numero = date('w', $temps);
$jours_complet = $jours[$jours_numero];
$NumeroDuJour = date('d', $temps);
$mois = array(' ', 'Janvier', 'Février', 'Mars', 'Avril', 'Mai', 'Juin', 'Juillet', 'Août', 'Septembre', 'Octobre', 'Novembre', 'Décembre');
$mois_numero = date("n", $temps);
$mois_complet = $mois[$mois_numero];
$an = date('Y', $temps);
$fr_temps = "$jours_complet, $NumeroDuJour $mois_complet $an";
return "$fr_temps";
}
echo "<br/>".f_date();
And i want to generate the following result:
<select name="ladate">
<option selected="selected" value="Mardi, 29 mai 2012">29 mai 2012</option>
<option value="Mercredi, 30 mai 2012">30 mai 2012</option></select>
....
<option value="Vendredi, 15 juin 2012">15 juin 2012</option></select>
</select>
Please tell me if you need further information.
Thank you.
Since you are only looking for MTWTF and you want the next 10 days, you can always safely look for the next 14 days and ignore the weekends and that will give 10 days. It won't work for holidays or anything like that but you can change it if you need to do that. I'll give you the pseudo-code here, I'll leave all the array mapping and text output to you
for ($days_to_add : 1 to 14) {
$new_date = date_add($days_to_add);
// check the day of the week
if (date('N', $new_date) >= 6) {
// ignore it, it's a weekend
continue;
}
// output the option tag for $new_date
echo "<option ... </option>"
}
This relies on the 10 and 14 day assumption, if you wanted to change that number you could add some sort of counter and only increment the counter if you were looking at a weekday/non-holiday
Just create a loop, increase the days up to ten and ignore all non open days (Sat, Sun). date('N') is your friend to detect the weekday of a given date.
<?php
$i = $openDay = 0;
while($openDay < 10) {
$i++;
$time = strtotime('+'.$i.' days');
$day = date('N', $time);
if ($day == 6 or $day == 7) { // ignore Saturdays and Sundays
continue;
}
echo f_date($time).'<br>';
$openDay++;
}
You also have to modify your date_f() function to use $temps as parameter.
<?php
function f_date($temps = null) {
// $temps = time();
// ...
}
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>';