the code snippet below shows a simulation of installments, with fixed expiration dates, divided every 30 days!
I would like to include a periodic expiration date, for example, every 20 days, or 10 every 10, depending on the variable $periodicity;
I have no idea how to do it?
<?php
function calculate_due($num_installment, $first_due_date = null){
if($first_due_date != null)
{
$first_due_date = explode('/',$first_due_date);
$day = $first_due_date[0];
$month = $first_due_date[1];
$year = $first_due_date[2];
}
else
{
$day = date('d');
$month = date('m');
$year = date('Y');
}
$periodicity = 20;
for($installment = 0; $installment < $num_installment; $installment++)
{
if ($periodicity == 30)
echo date('d/m/Y', strtotime('+'.$installment. " month", mktime(0, 0, 0, $month, $day, $year))),'<br/>';
else
echo date('d/m/Y', strtotime('+'.$installment. " month", mktime(0, 0, 0, $month, $periodicity, $year))),'<br/>';
}
}
echo 'Calculates installments from an informed date<br/>';
calculate_due(5, '10/10/2020');
Try this,
<?php
function calculate_due($num_installment, $first_due_date = null, $days = 1){
$start = DateTime::createFromFormat('d/m/Y', $first_due_date);
$end = DateTime::createFromFormat('d/m/Y', $first_due_date);
$end->add(new DateInterval('P'.($num_installment * $days).'D'));
$period = new DatePeriod(
$start,
new DateInterval('P'.$days.'D'),
$end
);
$return = [];
foreach ($period as $date) {
$return[] = $date->format('d/m/Y');
}
return $return;
}
echo 'Calculates installments from an informed date<br/>'.PHP_EOL;
echo implode("\n", calculate_due(5, '10/10/2020', 20));
https://3v4l.org/TLR8a
Change 20 (the last function parameter) to suit the number of days between.
Result:
Calculates installments from an informed date<br/>
10/10/2020<br/>
30/10/2020<br/>
19/11/2020<br/>
09/12/2020<br/>
29/12/2020<br/>
I have project built using laravel and a I have to build a function that counts all the complete quarters that are in the selected date range - the dates used are inserted via input.
Here are the quarters(i used numerical representations for the months)
01 - 03 first quarter
04 - 06 second quarter
07 - 09 third quarter
10 - 12 forth quarter
I would really appreciate your help,because I've been at it for an entire day now and basically have nothing to show for it,i thing I've been trying so hard i'm actually at the point where i'm so tired, i can t think straight.
I do have some code but it;s worthless, because it doesn't work, and any kind of idea or snippet of code is welcomed.
Thanks for your help in advance.
I managed to do this using multiple functions; basically, if this is needed for chart statistics, then a more specific approach might be the case.
I have done this in Laravel with timestamp dates as input (this code can be adapted for getting semesters also :) , it works and is already tested):
public static function getQuartersBetween($start_ts, $end_ts)
{
$quarters = [];
$months_per_year = [];
$years = self::getYearsBetween($start_ts, $end_ts);
$months = self::getMonthsBetween($start_ts, $end_ts);
foreach ($years as $year) {
foreach ($months as $month) {
if ($year->format('Y') == $month->format('Y')) {
$months_per_year[$year->format('Y')][] = $month;
}
}
}
foreach ($months_per_year as $year => $months) {
$january = new Date('01-01-' . $year);
$march = new Date('01-03-' . $year);
$april = new Date('01-04-' . $year);
$june = new Date('01-06-' . $year);
$july = new Date('01-07-' . $year);
$september = new Date('01-09-' . $year);
$october = new Date('01-10-' . $year);
$december = new Date('01-12-' . $year);
if (in_array($january, $months) && in_array($march, $months)) {
$quarter_per_year['label'] = 'T1 / ' . $year;
$quarter_per_year['start_day'] = $january->startOfMonth();
$quarter_per_year['end_day'] = $march->endOfMonth()->endOfDay();
array_push($quarters, $quarter_per_year);
}
if (in_array($april, $months) && in_array($june, $months)) {
$quarter_per_year['label'] = 'T2 / ' . $year;
$quarter_per_year['start_day'] = $april->startOfMonth();
$quarter_per_year['end_day'] = $june->endOfMonth()->endOfDay();
array_push($quarters, $quarter_per_year);
}
if (in_array($july, $months) && in_array($september, $months)) {
$quarter_per_year['label'] = 'T3 / ' . $year;
$quarter_per_year['start_day'] = $july->startOfMonth();
$quarter_per_year['end_day'] = $september->endOfMonth()->endOfDay();
array_push($quarters, $quarter_per_year);
}
if (in_array($october, $months) && in_array($december, $months)) {
$quarter_per_year['label'] = 'T4 / ' . $year;
$quarter_per_year['start_day'] = $october->startOfMonth();
$quarter_per_year['end_day'] = $december->endOfMonth()->endOfDay();
array_push($quarters, $quarter_per_year);
}
}
return $quarters;
}
and getting the years between:
public static function getYearsBetween($start_ts, $end_ts, $full_period = false)
{
$return_data = [];
$current = mktime(0, 0, 0, date('m', $start_ts), date('d', $start_ts), date('Y', $start_ts));
while ($current < $end_ts) {
$temp_date = $current;
$year = new Date($temp_date);
$return_data[] = $year;
$current = strtotime("+1 year", $current); // add a year
}
if ($full_period) {
$return_data[] = $end_ts;
}
return $return_data;
}
, also getting the months needed
public static function getMonthsBetween($start_ts, $end_ts, $full_period = false)
{
$return_data = $month_list = [];
$current = mktime(0, 0, 0, date('m', $start_ts), date('d', $start_ts), date('Y', $start_ts));
while ($current <= $end_ts) {
$temp_date = $current;
$date = new Date($temp_date);
$month_list[] = $date;
$current = strtotime("+1 month", $current); // add a month
}
$start_date_last_month = new Date(array_first($month_list));
$start_date_last_month = $start_date_last_month->startOfMonth()->format('m-d');
$temp_end_date = new Date($start_ts);
$temp_end_date = $temp_end_date->format('m-d');
if ($start_date_last_month < $temp_end_date) {
array_shift($month_list);
}
$end_date_last_month = new Date(end($month_list));
$current_day_month = $end_date_last_month->endOfMonth()->format('m-d');
$temp_end_date = new Date($end_ts);
$end_day_of_month = $temp_end_date->format('m-d');
if ($end_day_of_month < $current_day_month) {
array_pop($month_list);
}
if (count($month_list) == 0) {
$month_list[] = $end_date_last_month->subMonth();
}
$return_data = $month_list;
if ($full_period) {
$return_data[] = $end_ts;
}
return $return_data;
}
You can do something like in this example:
$February = 2;
$October = 10;
$completedQuarters = ceil($October/3) - ceil($February/3); // = 3
What about the quarter in which the date range starts, should it also count? If it should only count if it begins in the first month of a quarter you can check for it like this:
$completedQuarters = ceil($October/3) - ceil($February/3) -1; // = 2
if($February-1%3 == 0) $completedQuarters += 1;
You´re description is not very clear, let me know if that´s what you had in mind.
Not sure if the following is what you are meaning but might be useful
$date_start='2015/03/12';
$date_end='2017/11/14';
$timezone=new DateTimeZone('Europe/London');
$start=new DateTime( $date_start, $timezone );
$end=new DateTime( $date_end, $timezone );
$difference = $end->diff( $start );
$months = ( ( $difference->format('%y') * 12 ) + $difference->format('%m') );
$quarters = intval( $months / 3 );
printf( 'Quarters between %s and %s is %d covering %d months', $start->format('l, jS F Y'), $end->format('l, jS F Y'), $quarters, $months );
/*
This will output
----------------
Quarters between Thursday, 12th March 2015 and Tuesday, 14th November 2017 is 10 covering 32 months
*/
Something like this in the function and you should be set.
use Carbon\Carbon;
$first = Carbon::parse('2012-1-1'); //first param
$second = Carbon::parse('2014-9-15'); //second param
$fY = $first->year; //2012
$fQ = $first->quarter; //1
$sY = $second->year; //2014
$sQ = $second->quarter; //3
$n = 0; //the number of quarters we have counted
$i = 0; //an iterator we will use to determine if we are in the first year
for ($y=$fY; $y < $sY; $y++, $i++) { //for each year less than the second year (if any)
$s = ($i > 0) ? 1 : $fQ; //determine the starting quarter
for ($q=$s; $q <= 4; $q++) { //for each quarter
$n++; //count it
}
}
if ($sY > $fY) { //if both dates are not in the same year
$n = $n + $sQ; //total is the number of quarters we've counted plus the second quarter value
} else {
for ($q=$fQ; $q <= $sQ; $q++) { //for each quarter between the first quarter and second
$n++; //count it
}
}
print $n; //the value to return (11)
I have an array of random dates (not coming from MySQL). I need to group them by the week as Week1, Week2, and so on upto Week5.
What I have is this:
$dates = array('2015-09-01','2015-09-05','2015-09-06','2015-09-15','2015-09-17');
What I need is a function to get the week number of the month by providing the date.
I know that I can get the weeknumber by doing
date('W',strtotime('2015-09-01'));
but this week number is the number between year (1-52) but I need the week number of the month only, e.g. in Sep 2015 there are 5 weeks:
Week1 = 1st to 5th
Week2 = 6th to 12th
Week3 = 13th to 19th
Week4 = 20th to 26th
Week5 = 27th to 30th
I should be able to get the week Week1 by just providing the date
e.g.
$weekNumber = getWeekNumber('2015-09-01') //output 1;
$weekNumber = getWeekNumber('2015-09-17') //output 3;
I think this relationship should be true and come in handy:
Week of the month = Week of the year - Week of the year of first day of month + 1
We also need to make sure that "overlapping" weeks from the previous year are handeled correctly - if January 1st is in week 52 or 53, it should be counted as week 0. In a similar fashion, if a day in December is in the first week of the next year, it should be counted as 53. (Previous versions of this answer failed to do this properly.)
<?php
function weekOfMonth($date) {
//Get the first day of the month.
$firstOfMonth = strtotime(date("Y-m-01", $date));
//Apply above formula.
return weekOfYear($date) - weekOfYear($firstOfMonth) + 1;
}
function weekOfYear($date) {
$weekOfYear = intval(date("W", $date));
if (date('n', $date) == "1" && $weekOfYear > 51) {
// It's the last week of the previos year.
return 0;
}
else if (date('n', $date) == "12" && $weekOfYear == 1) {
// It's the first week of the next year.
return 53;
}
else {
// It's a "normal" week.
return $weekOfYear;
}
}
// A few test cases.
echo weekOfMonth(strtotime("2020-04-12")) . " "; // 2
echo weekOfMonth(strtotime("2020-12-31")) . " "; // 5
echo weekOfMonth(strtotime("2020-01-02")) . " "; // 1
echo weekOfMonth(strtotime("2021-01-28")) . " "; // 5
echo weekOfMonth(strtotime("2018-12-31")) . " "; // 6
To get weeks that starts with sunday, simply replace date("W", ...) with strftime("%U", ...).
You can use the function below, fully commented:
/**
* Returns the number of week in a month for the specified date.
*
* #param string $date
* #return int
*/
function weekOfMonth($date) {
// estract date parts
list($y, $m, $d) = explode('-', date('Y-m-d', strtotime($date)));
// current week, min 1
$w = 1;
// for each day since the start of the month
for ($i = 1; $i < $d; ++$i) {
// if that day was a sunday and is not the first day of month
if ($i > 1 && date('w', strtotime("$y-$m-$i")) == 0) {
// increment current week
++$w;
}
}
// now return
return $w;
}
The corect way is
function weekOfMonth($date) {
$firstOfMonth = date("Y-m-01", strtotime($date));
return intval(date("W", strtotime($date))) - intval(date("W", strtotime($firstOfMonth)));
}
I have created this function on my own, which seems to work correctly. In case somebody else have a better way of doing this, please share.. Here is what I have done.
function weekOfMonth($qDate) {
$dt = strtotime($qDate);
$day = date('j',$dt);
$month = date('m',$dt);
$year = date('Y',$dt);
$totalDays = date('t',$dt);
$weekCnt = 1;
$retWeek = 0;
for($i=1;$i<=$totalDays;$i++) {
$curDay = date("N", mktime(0,0,0,$month,$i,$year));
if($curDay==7) {
if($i==$day) {
$retWeek = $weekCnt+1;
}
$weekCnt++;
} else {
if($i==$day) {
$retWeek = $weekCnt;
}
}
}
return $retWeek;
}
echo weekOfMonth('2015-09-08') // gives me 2;
function getWeekOfMonth(DateTime $date) {
$firstDayOfMonth = new DateTime($date->format('Y-m-1'));
return ceil(($firstDayOfMonth->format('N') + $date->format('j') - 1) / 7);
}
Goendg solution does not work for 2016-10-31.
function weekOfMonth($strDate) {
$dateArray = explode("-", $strDate);
$date = new DateTime();
$date->setDate($dateArray[0], $dateArray[1], $dateArray[2]);
return floor((date_format($date, 'j') - 1) / 7) + 1;
}
weekOfMonth ('2015-09-17') // returns 3
Given the time_t wday (0=Sunday through 6=Saturday) of the first of the month in firstWday, this returns the (Sunday-based) week number within the month:
weekOfMonth = floor((dayOfMonth + firstWday - 1)/7) + 1
Translated into PHP:
function weekOfMonth($dateString) {
list($year, $month, $mday) = explode("-", $dateString);
$firstWday = date("w",strtotime("$year-$month-1"));
return floor(($mday + $firstWday - 1)/7) + 1;
}
You can also use this simple formula for finding week of the month
$currentWeek = ceil((date("d",strtotime($today_date)) - date("w",strtotime($today_date)) - 1) / 7) + 1;
ALGORITHM :
Date = '2018-08-08' => Y-m-d
Find out day of the month eg. 08
Find out Numeric representation of the day of the week minus 1 (number of days in week) eg. (3-1)
Take difference and store in result
Subtract 1 from result
Divide it by 7 to result and ceil the value of result
Add 1 to result eg. ceil(( 08 - 3 ) - 1 ) / 7) + 1 = 2
My function. The main idea: we would count amount of weeks passed from the month's first date to current. And the current week number would be the next one. Works on rule: "Week starts from monday" (for sunday-based type we need to transform the increasing algorithm)
function GetWeekNumberOfMonth ($date){
echo $date -> format('d.m.Y');
//define current year, month and day in numeric
$_year = $date -> format('Y');
$_month = $date -> format('n');
$_day = $date -> format('j');
$_week = 0; //count of weeks passed
for ($i = 1; $i < $_day; $i++){
echo "\n\n-->";
$_newDate = mktime(0,0,1, $_month, $i, $_year);
echo "\n";
echo date("d.m.Y", $_newDate);
echo "-->";
echo date("N", $_newDate);
//on sunday increasing weeks passed count
if (date("N", $_newDate) == 7){
echo "New week";
$_week += 1;
}
}
return $_week + 1; // as we are counting only passed weeks the current one would be on one higher
}
$date = new DateTime("2019-04-08");
echo "\n\nResult: ". GetWeekNumberOfMonth($date);
$month = 6;
$year = 2021;
$week = date("W", strtotime($year . "-" . $month ."-01"));
$str='';
$str .= date("d-m-Y", strtotime($year . "-" . $month ."-01")) ."to";
$unix = strtotime($year."W".$week ."+1 week");
while(date("m", $unix) == $month){
$str .= date("d-m-Y", $unix-86400) . "|";
$str .= date("d-m-Y", $unix) ."to";
$unix = $unix + (86400*7);
}
$str .= date("d-m-Y", strtotime("last day of ".$year . "-" . $month));
$weeks_ar = explode('|',$str);
echo '<pre>'; print_r($weeks_ar);
working fine.
// Current week of the month starts with Sunday
$first_day_of_the_week = 'Sunday';
$start_of_the_week1 = strtotime("Last $first_day_of_the_week");
if (strtolower(date('l')) === strtolower($first_day_of_the_week)) {
$start_of_the_week1 = strtotime('today');
}
$end_of_the_week1 = $start_of_the_week1 + (60 * 60 * 24 * 7) - 1;
// Get the date format
print date('Y-m-d', $start_of_the_week1) . ' 00:00:00';
print date('Y-m-d', $end_of_the_week1) . ' 23:59:59';
// self::DAYS_IN_WEEK = 7;
function getWeeksNumberOfMonth(): int
{
$currentDate = new \DateTime();
$dayNumberInMonth = (int) $currentDate->format('j');
$dayNumberInWeek = (int) $currentDate->format('N');
$dayNumberToLastSunday = $dayNumberInMonth - $dayNumberInWeek;
$daysCountInFirstWeek = $dayNumberToLastSunday % self::DAYS_IN_WEEK;
$weeksCountToLastSunday = ($dayNumberToLastSunday - $daysCountInFirstWeek) / self::DAYS_IN_WEEK;
$weeks = [];
array_push($weeks, $daysCountInFirstWeek);
for ($i = 0; $i < $weeksCountToLastSunday; $i++) {
array_push($weeks, self::DAYS_IN_WEEK);
}
array_push($weeks, $dayNumberInWeek);
if (array_sum($weeks) !== $dayNumberInMonth) {
throw new Exception('Logic is not valid');
}
return count($weeks);
}
Short variant:
(int) (new \DateTime())->format('W') - (int) (new \DateTime('first day of this month'))->format('W') + 1;
There is a many solutions but here is one my solution that working well in the most cases.
function current_week ($date = NULL) {
if($date) {
if(is_numeric($date) && ctype_digit($date) && strtotime(date('Y-m-d H:i:s',$date)) === (int)$date)
$unix_timestamp = $date;
else
$unix_timestamp = strtotime($date);
} else $unix_timestamp = time();
return (ceil((date('d', $unix_timestamp) - date('w', $unix_timestamp) - 1) / 7) + 1);
}
It accept unix timestamp, normal date or return current week from the time() if you not pass any value.
Enjoy!
I know this an old post but i have an idea!
$datetime0 = date_create("1970-01-01");
$datetime1 = date_create(date("Y-m-d",mktime(0,0,0,$m,"01",$Y)));
$datetime2 = date_create(date("Y-m-d",mktime(0,0,0,$m,$d,$Y)));
$interval1 = date_diff($datetime0, $datetime1);
$daysdiff1= $interval1->format('%a');
$interval2 = date_diff($datetime0, $datetime2);
$daysdiff2= $interval2->format('%a');
$week1=round($daysdiff1/7);
$week2=round($daysdiff2/7);
$WeekOfMonth=$week2-$week1+1;
$date = new DateTime('first Monday of this month');
$thisMonth = $date->format('m');
$mondays_arr = [];
// Get all the Mondays in the current month and store in array
while ($date->format('m') === $thisMonth) {
//echo $date->format('Y-m-d'), "\n";
$mondays_arr[] = $date->format('d');
$date->modify('next Monday');
}
// Get the day of the week (1-7 from monday to sunday)
$day_of_week = date('N') - 1;
// Get the day of month (1 to 31)
$current_week_monday_date = date('j') - $day_of_week;
/*$day_of_week = date('N',mktime(0, 0, 0, 2, 11, 2020)) - 1;
$current_week_monday_date = date('j',mktime(0, 0, 0, 2, 11, 2020)) - $day_of_week;*/
$week_no = array_search($current_week_monday_date,$mondays_arr) + 1;
echo "Week No: ". $week_no;
How about this function making use of PHP's relative dates?
This function assumes the week ends on Saturday. But this can be changed easily.
function get_weekNumMonth($date) {
$CI = &get_instance();
$strtotimedate = strtotime($date);
$firstweekEnd = date('j', strtotime("FIRST SATURDAY OF " . date("F", $strtotimedate) . " " . date("Y", $strtotimedate)));
$cutoff = date('j', strtotime($date));
$weekcount = 1;
while ($cutoff > $firstweekEnd) {
$weekcount++;
$firstweekEnd += 7; // move to next week
}
return $weekcount;
}
This function returns the integer week number of the current month. Weeks always start on Monday and counting always starts with 1.
function weekOfmonth(DateTime $date)
{
$dayFirstMonday = date_create('first monday of '.$date->format('F Y'))->format('j');
return (int)(($date->format('j') - $dayFirstMonday +7)/7) + ($dayFirstMonday == 1 ? 0 : 1);
}
Example of use
echo weekOfmonth(new DateTime("2020-04-12")); //2
A test for all days from 1900-2038 with the accepted solution from #Anders as a reference:
//reference functions
//integer $date (Timestamp)
function weekOfMonthAnders($date) {
//Get the first day of the month.
$firstOfMonth = strtotime(date("Y-m-01", $date));
//Apply above formula.
return weekOfYear($date) - weekOfYear($firstOfMonth) + 1;
}
function weekOfYear($date) {
$weekOfYear = intval(date("W", $date));
if (date('n', $date) == "1" && $weekOfYear > 51) {
// It's the last week of the previos year.
return 0;
}
else if (date('n', $date) == "12" && $weekOfYear == 1) {
// It's the first week of the next year.
return 53;
}
else {
// It's a "normal" week.
return $weekOfYear;
}
}
//this function
function weekOfmonth(DateTime $date)
{
$dayFirstMonday = date_create('first monday of '.$date->format('F Y'))->format('j');
return (int)(($date->format('j') - $dayFirstMonday +7)/7) + ($dayFirstMonday == 1 ? 0 : 1);
}
$dt = date_create('1900-01-01');
$end = date_create('2038-01-02');
$countOk = 0;
$countError = 0;
for(;$dt < $end; $dt->modify('+1 Day')){
$ts = $dt->getTimestamp();
if(weekOfmonth($dt) === weekOfMonthAnders($ts)){
++$countOk;
}
else {
++$countError;
}
}
echo $countOk.' compare ok, '.$countError.' errors';
Result: 50405 compare ok, 0 errors
I took the visual approach (like how we do it in the real world). Instead of using formulas or what not, I solved it (or at least I think I did) by visualizing a literal calendar and then putting the dates in a multidimensional array. The first dimension corresponds to the week.
I hope someone can check if it stands your tests. Or help someone out with a different approach.
# date in this format 2021-08-03
# week_start is either Sunday or Monday
function getWeekOfMonth($date, $week_start = "Sunday"){
list($year, $month, $day) = explode("-", $date);
$dates = array();
$current_week = 1;
$new_week_signal = $week_start == "Sunday" ? 6 : 0;
for($i = 1; $i <= date("t", strtotime($date)); $i++){
$current_date = strtotime("{$year}-{$month}-".$i);
$dates[$current_week][] = $i;
if(date('w', $current_date) == $new_week_signal){
$current_week++;
}
}
foreach($dates as $week => $days){
if(in_array(intval($day), $days)){
return $week;
}
}
return false;
}
//It's easy, no need to use php function
//Let's say your date is 2017-07-02
$Date = explode("-","2017-07-02");
$DateNo = $Date[2];
$WeekNo = $DateNo / 7; // devide it with 7
if(is_float($WeekNo) == true)
{
$WeekNo = ceil($WeekNo); //So answer will be 1
}
//If value is not float then ,you got your answer directly
I am working on a "cut-off" date and I need to set it on every month. Say, I set the cut-off date to August 25 for this year, then it should be September 25, October 25 and so on till the year ends.
Here's the code I have:
$now = "2015-08-25";
$nextDate = getCutoffDate($now);
echo $nextDate;
function getCutoffDate($start_date)
{
$date_array = explode("-",$start_date); // split the array
// var_dump($date_array);
$year = $date_array[0];
$month = $date_array[1];
$day = $date_array[2];
/*if (date('n', $now)==12)
{
return date("Y-m-d",strtotime(date("Y-m-d", $start_date) . "+1 month"));
}*/
if (date("d") <= $day) {
$billMonth = date_format(date_create(), 'm');
}
else{
$billMonth = date_format(date_modify(date_create(), 'first day of next month'), 'm');
}
// echo date("d").' '. $billMonth.'<br>';
$billMonthDays = cal_days_in_month(CAL_GREGORIAN, ($billMonth), date("Y"));
// echo $billMonthDays.'<br>';
if ($billMonthDays > $day) {
$billDay = $day;
} else {
$billDay = $billMonthDays;
}
}
I got this from here: http://www.midwesternmac.com/blogs/jeff-geerling/php-calculating-monthly
It returns the same date for the next month only, but how do I get the specific date of EACH month of the current year? Kindly leave your thoughts. Still a newbie here, sorry.
In that case, this should be enough:
<?php
for($i=8; $i<=12; $i++) {
echo sprintf('25-%02d-2015', $i) . '<br>';
}
but if you need more flexible way:
<?php
$date = new DateTime('25-08-2015');
function getCutoffDate($date) {
$days = cal_days_in_month(CAL_GREGORIAN, $date->format('n'), $date->format('Y'));
$date->add(new DateInterval('P' . $days .'D'));
return $date;
}
for($i = 0; $i < 5; $i++) {
$date = getCutoffDate($date);
echo $date->format('d-m-Y') . '<br>';
}
This should print:
25-09-2015
25-10-2015
25-11-2015
25-12-2015
25-01-2016
i want to calculate latest 31-Mar .... suppose date is 1-Jan-2012 i want result as 31-mar-2011 and if is 1-April-2011 then also i want result 31-mar-2011 and if its 1-mar-2011 it should come as 31-mar-2010.....hope i made my self clear ...(with php) ... i al calculating date with this for financial year ...
always between 31-mar-lastyear to 1-April-thisyear
... year should be taken automatically ...
i was trying like this
31-mar-date('y') and 31-mar-date('y')-1
but its not working as its taking current year every time.
Here is an example using the wonderful strtotime function of php.
<?php
$day = 1;
$month = 1;
$year = 2011;
$date = mktime(12, 0, 0, $month, $day, $year);
$targetMonth = 3;
$difference = $month - $targetMonth;
if($difference < 0) {
$difference += 12;
}
$sameDateInMarch = strtotime("- " . $difference . " months", $date);
echo "Entered date: " . date("d M Y", $date) . "<br />";
echo "Last 31 march: " . date("t M Y", $sameDateInMarch);
// the parameter 't' displays the last day of the month
?>
Something like this:
function getLast() {
$currentYear = date('Y');
// Check if it happened this year, AND it's not in the future.
$today = new DateTime();
if (checkdate(3, 31, $currentYear) && $today->getTimestamp() > mktime(0, 0, 0, 3, 31, $currentYear)) {
return $currentYear;
}
while (--$currentYear) {
if (checkdate(3, 31, $currentYear)) {
return $currentYear;
}
}
return false;
}
var_dump(getLast());
It should return the year or false.
if (date('m')>3) {
$year = date('Y').'-'.(date('Y')+1);
} else {
$year = (date('Y')-1).'-'.date('Y');
}
echo $year;
this is to get the current financial year for India