Get exact date by providing week number of month and day? - php

Is there any php function where i can get exact date, by providing week number of the month and day as,
function required_date($week_num, $day) {
// should return 20th of September, if i pass (3, 'Friday')
}
I am getting week number of the month as
ceil(substr(date('Y-m-d'), -2) / 7);
and day as,
date('l');
Any help, please!

function required_date($week_num, $day) {
$week_of_year = sprintf('%02d', date('W', strtotime(date('Y-m-01'))) + $week_num);
$day_of_week = date('N', strtotime($day));
$timestamp = strtotime(date('Y') . '-W' . $week_of_year . '-' . $day_of_week);
return $timestamp;
}
$timestamp = required_date(3, 'friday');
echo date('Y-m-d', $timestamp);

try this:
date('Y-m-d', strtotime('W' . $week_num . ' ' . $day));

How about
function required_date($week_num, $day)
{
return date('Y-m-') . substr('0' . (($week_num - 1) * 7 + date('N', strtotime($day))) + 1, -2);
}
var_dump(required_date(3, 'Friday'));
var_dump(required_date(4, 'Friday'));

Related

Calculating and adjusting dates based on exceptions (PHP)

Here are the rules:
The adding of days is always by 15s (ex. 15, 30, 45, 60, etc.)
When the maturity date day falls on the 15th or end of the month
(ex. 30 or 31 depends on the month, 28 or 29 every February depends
if leap year) when adding days (as mention above) the date
should fall ONLY to 15th or end of month.
When the maturity date day does not fall every 15th or end of the
month just normally add days.
When the date is February 14 and add 15 days it should return 02/29 if leap year or 02/28 if not leap year.
Here is my code but, I am getting 1 error and inconsistency.
Catchable fatal error when the date is 02/29/2020 and add 30 days.
What can I do to accommodate this rules?
function adjustDate($maturitydate, $add) {
$nodays = '+'.$add.' days';
$date = new DateTime($maturitydate);
$matdt = $date->modify($nodays);
$ismaturitydateendofmonth = check_end_of_month($maturitydate);
if($date->format('d') == 15) {
$matdt = $matdt->format('m/15/Y');
}
else if($ismaturitydateendofmonth == '1'){
$matdt->modify('last day of this month');
}
else{
$matdt = $matdt->format('m/d/Y');
}
return $matdt;
}
function check_end_of_month($date){
//adds 1 day to date
$Temp = date('m/d/Y',strtotime("+1 day", strtotime($date)));
//get the month of each date
$tempmonth = date('m', strtotime($Temp));
$datemonth = date('m', strtotime($date));
//check if the months are equal
if($tempmonth != $datemonth){
return '1';
}
else{
return '0';
}
}
The code below will fix all the exceptions and inconsistencies.
$matdt = $maturitydatefrom;
for ($x = 0; $x < $chknumdif; $x++) {
$matdt = getNextDuedate($matdt, $maturitydatefrom);
}
function getNextDuedate($prevdate, $maturitydate){
$maturityday = date('d', strtotime($maturitydate));
$prevday = date('d', strtotime($prevdate));
$prevmonth = date('m', strtotime($prevdate));
$prevyear = date('Y', strtotime($prevdate));
$prevlastday = date('t', strtotime($prevdate));
$maturitylastday = date('t', strtotime($maturitydate));
$isendofmonth = check_end_of_month($maturitydate);
if($maturityday == $maturitylastday || $maturityday == 15){
if($prevday == 15){
$duedate = $prevmonth . '/' . $prevlastday . '/' . $prevyear;
}
else{
$prevdate = date('m/d/Y', strtotime("+1 month", strtotime($prevdate)));
$duedate = date('m', strtotime($prevdate)) . '/15/' . date('Y', strtotime($prevdate));
}
}
else{
if($prevday < 15){
if($prevmonth == '2' && $prevday == '14'){
$duedate = $prevmonth . '/' . $prevlastday . '/' . $prevyear;
}
else{
$duedate = $prevmonth . '/' . ($prevday + 15) . '/' . $prevyear;
}
}
else if($prevday > 15){
if($maturityday < 15){
$prevdate = date('m/d/Y', strtotime("+1 month", strtotime($prevdate)));
$duedate = date('m', strtotime($prevdate)) . '/' . $maturityday . '/' . date('Y', strtotime($prevdate));
}
else{
$prevdate = date('m/d/Y', strtotime("+1 month", strtotime($prevdate)));
$duedate = date('m', strtotime($prevdate)) . '/' . ($maturityday - 15) . '/' . date('Y', strtotime($prevdate));
}
}
}
return $duedate;
}

How-to get the start/end Mondays of the month in php

I need to include also the last Monday of the precedent month, and the first Monday of the next month.
Example
2016-01-25
2016-02-01
2016-02-08
2016-02-15
2016-02-22
2016-02-29
2016-03-07
I have this code so far:
function getAllDaysInAMonth($year, $month, $day = 'Monday', $daysError = 3) {
$dateString = 'first ' . $day . ' of ' . $year . '-' . $month;
if (!strtotime($dateString)) {
throw new \Exception('"' . $dateString . '" is not a valid strtotime');
}
$startDay = new \DateTime($dateString);
if ($startDay->format('j') > $daysError) {
$startDay->modify('- 7 days');
}
$days = array();
while ($startDay->format('Y-m') <= $year . '-' . str_pad($month, 2, 0, STR_PAD_LEFT)) {
$days[] = clone($startDay);
$startDay->modify('+ 7 days');
}
return $days;
}
Ok, resolved!
function getAllDaysInAMonth($year, $month, $day = 'Monday', $daysError = 3) {
$dateString = 'first ' . $day . ' of ' . $year . '-' . $month;
if (!strtotime($dateString)) {
throw new \Exception('"' . $dateString . '" is not a valid strtotime');
}
$startDay = new \DateTime($dateString);
if ($startDay->format('j') > $daysError) {
$startDay->modify('- 7 days');
}
$days = array();
$lastMonday = new DateTime("last Monday of last month");
$nextMonday = new DateTime("first Monday of next month");
$days[] = clone($lastMonday);
while ($startDay->format('Y-m') <= $year . '-' . str_pad($month, 2, 0, STR_PAD_LEFT)) {
$days[] = clone($startDay);
$startDay->modify('+ 7 days');
}
$days[] = clone($nextMonday);
return $days;
}
Try this :
$lastMonday = new DateTime("last Monday of last month");
$nextMonday = new DateTime("first Monday of next month");
echo 'Last Monday : '.$lastMonday->format('Y-m-d').'<br />';
echo 'First Monday : '.$nextMonday->format('Y-m-d');

Get week number in month from date in PHP?

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

PHP - get nearest date for input of 'dd.mm'

When having a date string in the format dd.mm. without a year-part, what's the best way to get the most logical (nearest to today) dd.mm.yyyy string?
Examples:
nDate('06.06.') === '06.06.2014' // '06.06.2014' is nearer to '14.03.2014' then '06.06.2013'
nDate('10.12.') === '10.12.2013' // '10.12.2013' is nearer to '14.03.2014' then '10.12.'2014'
My current code:
function nDate($date) {
$tmpDate = DateTime::createFromFormat('d.m.Y', $date . date('Y', strtotime('-1 year')));
foreach ($years as $year) {
$tmpDate2 = DateTime::createFromFormat('d.m.Y', $date . $year);
if ($tmpDate2 instanceof DateTime && (!($tmpDate instanceof DateTime) || abs($tmpDate2->getTimeStamp() - time()) < abs($tmpDate->getTimeStamp() - time()))) {
$tmpDate = $tmpDate2;
}
}
}
For given month and day create full dates for previous, current and next year.
Choose the one of the first three that has least difference betweeen that date and current date.
You can do that with strtotime function.
Edit:
It could be like this:
<?php
/**
* #param string $srcDate format mm-dd
*/
function closestDate($srcDate) {
$dates['prev'] = (date('Y') - 1) . '-' . $srcDate;
$dates['curr'] = date('Y') . '-' . $srcDate;
$dates['next'] = (date('Y') + 1) . '-' . $srcDate;
$times = array_map('strtotime', $dates);
$diffs = array_map(function($time) {
return abs($time - strtotime('today'));
}, $times);
$minDiff = min($diffs);
$key = array_keys($diffs, $minDiff);
return $dates[$key[0]];
}
echo closestDate('06-06'); //2014-06-06
echo closestDate('12-10'); //2013-12-10

PHP to calculate latest 31 March

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

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