I have date time data in database and from that im calculating total hours only taking the date not the timing.
im using strtotime for that calculations. For ex:
$LastDate = '2016-04-27 17:27:28';
$endDate = strtotime($LastDate); // 1461770848
$LastDate = '2016-04-27';
$endDate = strtotime($LastDate); // 1461708000
I want this second value only.
So i want to extract the date from the $LastDate .
///////////Getting data from Employee Outstation(Travel) details from database/////////////
$employeeTravel = new EmployeeTravelRecord();
$TravelEntryList = $employeeTravel->Find("employee = 2 and (date(travel_date) <= ? and ? <= date(return_date))",array($req['date_end'],$req['date_start']));
$startdate = $req['date_start'];
$enddate = $req['date_end'];
foreach($TravelEntryList as $Travelentry){
$key = $Travelentry->employee;
if($startdate >= $Travelentry->travel_date)
{
$firstdate = $startdate;
}
else
$firstdate = $Travelentry->travel_date;
if($enddate <= $Travelentry->return_date )
{
$lastdate = $enddate;
}
else
$lastdate = $Travelentry->return_date;
$holidays = $this->getholidays($firstdate,$lastdate);
$totalhours = $this->getWorkingDays($firstdate,$lastdate,$holidays);
$amount = $totalhours;
}
private function getWorkingDays($FirstDate,$LastDate,$holidays){
$endDate = strtotime($LastDate); //Here i want to give only date not the time
$startDate = strtotime($FirstDate);
$days = ($endDate - $startDate) / 86400 + 1;
$no_full_weeks = floor($days / 7);
$no_remaining_days = fmod($days, 7);
//Im doing even and odd saturday calculations....finally...
$workinghours = $workingDays*9 + $no_saturdays*7;
return $workinghours;
}
such expression gives you desired result
echo strtotime('midnight', strtotime('2016-04-27 22:11')); // 1461715200
echo strtotime('2016-04-27'); // 1461715200
Related
Anyone can help me to get all public holidays date between 2 dates?
I have a code to count the number of working days then I'm here asking for help on how to
get all public holidays between 2 date (the $given_date and $target_date) and hold it in array for me to replace the hard coded value of $holiday variable.
<?php
date_default_timezone_set("Asia/Manila");
function getWorkingDays($startDate, $endDate, $offdays, $holidays)
{
$endDate = strtotime($endDate);
$startDate = strtotime($startDate);
$days = ($endDate - $startDate) / 86400 + 1;
$counter = 0;
for($i = 1; $i <= $days; $i++)
{
$the_first_day_of_week = date("N", $startDate);
$startDate += 86400;
if(!in_array($the_first_day_of_week, $offdays) && !in_array(date("Y-m-d",$startDate), $holidays))
{
$counter++;
}
}
return $counter;
}
$given_date = "2021-10-20";
$target_date = "2021-11-06";
$holidays = array("2021-11-01", "2021-11-02");
$offdays = array(5,6);
echo getWorkingDays($given_date, $target_date, $offdays, $holidays);
?>
I have this code that modify date in the way I want, for example, if starting date is 31/01/2000, adding 1 month will return 29/02/2000, then, 31/03/2000.
If date is 30/01/2000 (not last day of the month) it will return 29/02/2000, then 30/03/2000, 30/04/2000 and so on.
I want to transfirm that code in a general case function, to be able to add 1,3,6,12 months, and inside the for loop, to work all corect.
I would like it to be a function 2 or 3 arguments, startingDate, duration(nr of iterations), frequency (add 1/3/6/12 months per once).
<?php
$date = new DateTime('2000-01-28'); // or whatever
#echo $date->format('d')." ".$date->format('t');
$expectedDay = $date->format('d');
$month = $date->format('m');
$year = $date->format('Y');
for ($i = 0; $i < 100; $i++) {
echo $date->format('Y-m-d') . "<br>";
if ($month++ == 12) {
$year++;
$month = 1;
}
$date->modify("${year}-${month}-1");
if ($expectedDay > $date->format('t')) {
$day = $date->format('t');
} else {
$day = $expectedDay;
}
$date->modify("${year}-${month}-${day}");
}
Whelp, there's an extremely easy function for this in PHP nowadays.
first, you get a timestamp instead of a datetime:
$timestamp = $date->getTimestamp();
Now, just use strtotime to add onto the date.
strtotime("+1 month", $myTimestamp);
You can change the +1 into anything you want, so you just throw the amount in the string and voila; a dynamic way of adding them!
however, since you want to do +30 days instead of a natural month, you're better off just adding 30 days to the timestamp, like so:
$timestamp = $timestamp + (3600 * 24 * 30); //s per h * h per d * d
So, you'd end up with something like this:
function calculateTime($startingDate, $iterations, $frequency){
$timeStamp = strtotime($startingDate);//if you expect a string date
$timeToAdd = (3600 * 24 * 30) * $frequency; //30 days * frequency
$return = array();
$return[] = date('Y-m-d', $timeStamp); //Original date
$previousDate = $timeStamp; //Original date for now
for($i = 0; $i < $iterations; $i++){
$newDate = $previousDate + (3600 * 24 * 30);
$return[] = date('Y-m-d', $newDate);
$previousDate = $newDate;
}
return $return;
}
And then for the rendering part:
//Let's render this stuff
$dates = calculateTime('24-08-2017', 25, 3);
foreach($dates as $date){
echo "$date</br>";
}
If you'd like to do it with full months, something like this:
<?php
function calculateTime($startingDate, $iterations, $frequency){
$timeStamp = strtotime($startingDate);//if you expect a string date
$return = array();
$return[] = date('Y-m-d', $timeStamp); //Original date
$previousDate = $timeStamp; //Original date for now
for($i = 0; $i < $iterations; $i++){
$lastDay = false;
//It's the last day of the month
if(date('t', $timeStamp) == date('d', $timeStamp)){
$lastDay = true;
}
if($frequency == 12){
$newDate = strtotime('+1 year', $previousDate);
}
else{
if($lastDay){
$firstDayOfMonth = strtotime(date("01-m-Y", $previousDate));
$newDate =strtotime("+$frequency month", $firstDayOfMonth);
}
else{
$newDate = strtotime("+$frequency month", $previousDate);
}
}
if($lastDay){
$return[] = date('Y-m-t', $newDate);
}
else{
$return[] = date('Y-m-d', $newDate);
}
$previousDate = $newDate;
}
return $return;
}
//Let's render this stuff
$dates = calculateTime('31-01-2000', 25, 1);
foreach($dates as $date){
echo "$date</br>";
}
I hope this helps? :)
If you'd like to see how this works quickly, just paste my code into a phpfiddle. Unfortunately the save function is broken right now.
I try this
<?php
$startdate = '2016-07-15';
$enddate = '2016-07-17';
$sundays = [];
$startweek=date("W",strtotime($startdate));
$endweek=date("W",strtotime($enddate));
$year=date("Y",strtotime($startdate));
for($i=$startweek;$i<=$endweek;$i++) {
$result=$this->getWeek($i,$year);
if($result>$startdate && $result<$enddate) {
$sundays[] = $result;
}
}
print_r($sundays);
public function getWeek($week, $year)
{
$dto = new \DateTime();
$result = $dto->setISODate($year, $week, 0)->format('Y-m-d');
return $result;
}
?>
this return blank array. but in between two dates 2016-07-17 is Sunday.
I get output as 2016-07-17
I refer this here
But in this link return output as no of sunday not date.
Give this a try:
$startDate = new DateTime('2016-07-15');
$endDate = new DateTime('2016-07-17');
$sundays = array();
while ($startDate <= $endDate) {
if ($startDate->format('w') == 0) {
$sundays[] = $startDate->format('Y-m-d');
}
$startDate->modify('+1 day');
}
var_dump($sundays);
If you want later to use the DateTime objects instead of the formatted date, then you must use DateTimeImmutable for the $startDate variable:
$startDate = new DateTimeImmutable('2016-07-15');
$endDate = new DateTimeImmutable('2016-07-17');
$sundays = array();
while ($startDate <= $endDate) {
if ($startDate->format('w') == 0) {
$sundays[] = $startDate;
}
$startDate->modify('+1 day');
}
var_dump($sundays);
function getDateForSpecificDayBetweenDates($startDate, $endDate, $weekdayNumber)
{
$startDate = strtotime($startDate);
$endDate = strtotime($endDate);
$dateArr = array();
do
{
if(date("w", $startDate) != $weekdayNumber)
{
$startDate += (24 * 3600); // add 1 day
}
} while(date("w", $startDate) != $weekdayNumber);
while($startDate <= $endDate)
{
$dateArr[] = date('Y-m-d', $startDate);
$startDate += (7 * 24 * 3600); // add 7 days
}
return($dateArr);
}
$dateArr = getDateForSpecificDayBetweenDates('2010-01-01', '2010-12-31', 0);
print "<pre>";
print_r($dateArr);
Try out this code..
Try this
$start = new DateTime($startDate);
$end = new DateTime($endDate);
$sundays = [];
while ($start->getTimestamp() != $end->getTimestamp()) {
if ($start->format('w') == 0) {
$sundays[] = $start->format('Y-m-d');
}
$start->add('+1 DAY');
}
This will return you all sundays between two dates.
$startdate = '2016-05-1';
$enddate = '2016-05-20';
function getSundays($start, $end) {
$timestamp1 = strtotime($start);
$timestamp2 = strtotime($end);
$sundays = array();
$oneDay = 60*60*24;
for($i = $timestamp1; $i <= $timestamp2; $i += $oneDay) {
$day = date('N', $i);
// If sunday
if($day == 7) {
// Save sunday in format YYYY-MM-DD, if you need just timestamp
// save only $i
$sundays[] = date('Y-m-d', $i);
// Since we know it is sunday, we can simply skip
// next 6 days so we get right to next sunday
$i += 6 * $oneDay;
}
}
return $sundays;
}
var_dump(getSundays($startdate, $enddate));
Use Carbon
$arrayOfDate = [];
$startDate = Carbon::parse($startDate)->modify('this sunday');
$endDate = Carbon::parse($endDate);
for ($date = $startDate; $date->lte($endDate); $date->addWeek()) {
$arrayOfDate[] = $date->format('Y-m-d');
}
return $arrayOfDate;
Let's say that I have two dates:
$initialDate = '08/10/2015 09:30:24 am';
$finalDate = '15/10/2015 15:47:38 pm';
$holiday = '12/10/2015';
I have to consider the hour of these days.
Hours to consider : 8 hours per day;
Start : 8 pm
End: 18 pm (24 hours format )
Lunch break start: 12:00 pm
Lunch break end: 14:00 pm
Example 1 : From 08/10/2015 10:00:00 to 09/10/2015 17:00:00 results 13 working hours. ( excludes lunch break )
Example 2 : From 08/10/2015 14:00:00 to 09/10/2015 18:00:00 results 12 working hours. ( Do not exclude 2 hours from begin date, because starts after 14:00 pm, lunch break )
Example 3 : From 08/10/2015 16:00:00 to 09/10/2015 18:00:00 results 10 working hours. ( Do not exclude 2 hours from begin date, because starts after 14:00 pmm lunch break )
Exampld 4 : From 08/10/2015 08:00:00 to 09/10/2015 11:00:00 results 14 working hours. ( Exclude 2 hours from begin date, and do not exclude 2 hours from end date, because isn't after 14:00 pm )
And I have to calculate the working hours and working days between those two dates, excluding weekends and Holidays, how can I do that ? I'm using PHP.
PS: I Already have something, but without lunch break... I made a research here on StackOverFlow.
Code:
function get_workdays($dataInicial,$dataFinal){
// arrays
$days_array = array();
$skipdays = array("Saturday", "Sunday");
$skipdates = get_feriados();
// other variables
$i = 0;
$current = $dataInicial;
if($current == $dataFinal) // same dates
{
$timestamp = strtotime($dataInicial);
if (!in_array(date("l", $timestamp), $skipdays)&&!in_array(date("Y-m-d", $timestamp), $skipdates)) {
$days_array[] = date("Y-m-d",$timestamp);
}
}
elseif($current < $dataFinal) // different dates
{
while ($current < $dataFinal) {
$timestamp = strtotime($dataInicial." +".$i." day");
if (!in_array(date("l", $timestamp), $skipdays)&&!in_array(date("Y-m-d", $timestamp), $skipdates)) {
$days_array[] = date("Y-m-d",$timestamp);
}
$current = date("Y-m-d",$timestamp);
$i++;
}
}
return $days_array;
}
function get_feriados(){
$dateAno = Date('Y');
$days_array = array(
$dateAno.'-10-12', // Padroeira do Brasil/ Dias das Crianças
$dateAno.'-11-02', // Finados
$dateAno.'-12-25' // Finados
);
return $days_array;
}
date_default_timezone_set('America/Sao_Paulo');
$dateAno = Date('Y');
$dataInicial = Date('08/10/2015 H:i');
$dataFinal = Date('13/10/2015 H:i');
// timestamps
$from_timestamp = strtotime(str_replace('/', '-', $dataInicial));
$to_timestamp = strtotime(str_replace('/', '-', $dataFinal));
// work day seconds
$workday_start_hour = 9;
$workday_end_hour = 17;
$workday_seconds = ($workday_end_hour - $workday_start_hour)*3600;
// work days beetwen dates, minus 1 day
$from_date = date('Y-m-d',$from_timestamp);
$to_date = date('Y-m-d',$to_timestamp);
$workdays_number = count(get_workdays($from_date,$to_date))-1;
$workdays_number = $workdays_number<0 ? 0 : $workdays_number;
// start and end time
$start_time_in_seconds = date("H",$from_timestamp)*3600+date("i",$from_timestamp)*60;
$end_time_in_seconds = date("H",$to_timestamp)*3600+date("i",$to_timestamp)*60;
// final calculations
$working_hours = ($workdays_number * $workday_seconds + $end_time_in_seconds - $start_time_in_seconds) / 86400 * 24;
print_r('<br/> Horas úteis '.$working_hours);
}
But don't consider two hours of break lunch. Can somebody please help me ?
If you use PHP 5.3 or higher, you can do this:
$datefrom = DateTime::createFromFormat('d/m/Y', '08/10/2015');
$dateto = DateTime::createFromFormat('d/m/Y', '15/10/2015');
$interval = $datefrom->diff($dateto);
$days = intval($interval->format('%a'));
Also you can remove holidays with if:
if ($datetime1->getTimestamp() < $holiday->getTimestamp() and $datetime2->getTimestamp() > $holiday->getTimestamp()) $days--;
Calculate hours between two days:
$datefrom = DateTime::createFromFormat('d/m/Y H:i:s', '08/10/2015 12:51:34');
$dateto = DateTime::createFromFormat('d/m/Y H:i:s', '15/10/2015 13:14:56');
$hours = intval($interval->format('%a')) * 24 + $interval->format('%h');
You can calculate hours of launches sum and then subtract it.
How to ignore weekends or calculate ignore days:
while($dateto->getTimestamp() > $datefrom->getTimestamp()) {
if (in_array($datefrom->format('w'), array('0','6'))) $ignore_days += 1;
$datefrom->modify('+1 day');
}
I expect this will do all you want. But I changed the datetime format as follows. Check it. Used less comments. If any query, please ask. Holidays are arrays, add and remove as required.
Times between 12:00 - 14:00 is handled.
Times below 08:00 is handled.
Times above 18:00 is handled.
<?php
$initialDate = '2015-10-13 08:15:00'; //start date and time in YMD format
$finalDate = '2015-10-14 11:00:00'; //end date and time in YMD format
$holiday = array('2015-10-12'); //holidays as array
$noofholiday = sizeof($holiday); //no of total holidays
//create all required date time objects
$firstdate = DateTime::createFromFormat('Y-m-d H:i:s',$initialDate);
$lastdate = DateTime::createFromFormat('Y-m-d H:i:s',$finalDate);
if($lastdate > $firstdate)
{
$first = $firstdate->format('Y-m-d');
$first = DateTime::createFromFormat('Y-m-d H:i:s',$first." 00:00:00" );
$last = $lastdate->format('Y-m-d');
$last = DateTime::createFromFormat('Y-m-d H:i:s',$last." 23:59:59" );
$workhours = 0; //working hours
for ($i = $first;$i<=$last;$i->modify('+1 day') )
{
$holiday = false;
for($k=0;$k<$noofholiday;$k++) //excluding holidays
{
if($i == $holiday[$k])
{
$holiday = true;
break;
} }
$day = $i->format('l');
if($day === 'Saturday' || $day === 'Sunday') //excluding saturday, sunday
$holiday = true;
if(!$holiday)
{
$ii = $i ->format('Y-m-d');
$f = $firstdate->format('Y-m-d');
$l = $lastdate->format('Y-m-d');
if($l ==$f )
$workhours +=sameday($firstdate,$lastdate);
else if( $ii===$f)
$workhours +=firstday($firstdate);
else if ($l ===$ii)
$workhours +=lastday($lastdate);
else
$workhours +=8;
}
}
echo $workhours; //echo the hours
}
else
echo "lastdate less than first date";
function sameday($firstdate,$lastdate)
{
$fmin = $firstdate->format('i');
$fhour = $firstdate->format('H');
$lmin = $lastdate->format('i');
$lhour = $lastdate->format('H');
if($fhour >=12 && $fhour <14)
$fhour = 14;
if($fhour <8)
$fhour =8;
if($fhour >=18)
$fhour =18;
if($lhour<8)
$lhour=8;
if($lhour>=12 && $lhour<14)
$lhour = 14;
if($lhour>=18)
$lhour = 18;
if($lmin == 0)
$min = ((60-$fmin)/60)-1;
else
$min = ($lmin-$fmin)/60;
return $lhour-$fhour + $min;
}
function firstday($firstdate) //calculation of hours of first day
{
$stmin = $firstdate->format('i');
$sthour = $firstdate->format('H');
if($sthour<8) //time before morning 8
$lochour = 8;
else if($sthour>18)
$lochour = 0;
else if($sthour >=12 && $sthour<14)
$lochour = 4;
else
{
$lochour = 18-$sthour;
if($sthour<=14)
$lochour-=2;
if($stmin == 0)
$locmin =0;
else
$locmin = 1-( (60-$stmin)/60); //in hours
$lochour -= $locmin;
}
return $lochour;
}
function lastday($lastdate) //calculation of hours of last day
{
$stmin = $lastdate->format('i');
$sthour = $lastdate->format('H');
if($sthour>=18) //time after 18
$lochour = 8;
else if($sthour<8) //time before morning 8
$lochour = 0;
else if($sthour >=12 && $sthour<14)
$lochour = 4;
else
{
$lochour = $sthour - 8;
$locmin = $stmin/60; //in hours
if($sthour>14)
$lochour-=2;
$lochour += $locmin;
}
return $lochour;
}
?>
Check the bellow code, that will return the number of Working days
function number_of_working_days($from, $to) {
$workingDays = [1, 2, 3, 4, 5];// date format = (1 = Monday,2 = Tue, ...)
$holidayDays = ['*-12-25', '*-02-14', '2015-12-23']; // variable and fixed holidays
$from = new DateTime($from);
$to = new DateTime($to);
$to->modify('+1 day');
$interval = new DateInterval('P1D');
$days = new DatePeriod($from, $interval, $to);
$no_of_working_days = 0;
foreach ($days as $day) {
if (!in_array($day->format('N'), $workingDays)||in_array($day->format('Y-m-d'), $holidayDays)||in_array($day->format('*-m-d'), $holidayDays)) {continue;}
$working_days++;
}
return $no_of_working_days;
}
echo number_of_working_days('2015-12-01', '2015-09-10');
From that you can easily calculate the Number of Working Hours.
I have created for you this nice class you can use. It requires the nesbot/carbon library (http://carbon.nesbot.com/) and you use it like so:
$calc = new HoursCalculator(
Carbon::createFromFormat("Y-m-d H:i", "2015-10-7 09:00"),
Carbon::createFromFormat("Y-m-d H:i", "2015-10-14 18:00"),
[
"2015-10-13"
]
);
echo $calc->getHours();
Heres the class:
class HoursCalculator {
const LUNCH_HOURS = 2;
protected $start;
protected $end;
protected $holidays;
protected $hoursTotal;
public function __construct(Carbon $start, Carbon $end, $holidays = [])
{
$this->start = $start;
$this->end = $end;
$this->holidays = $holidays;
}
public function getHours()
{
$dayHours = $this->getHoursInADay();
return $this->calculateHours($dayHours);
}
protected function getHoursInADay()
{
$start = $this->start;
$end = Carbon::createFromFormat("Y-m-d H:i", $this->start->format("Y-m-d") . " " . $this->end->format("H:i"));
return $start->diffInHours($end) - self::LUNCH_HOURS;
}
protected function getStartDate()
{
return $this->start->format('Y-m-d');
}
protected function calculateHours($hoursInDay)
{
$start = $this->start->copy()->startOfDay();
$end = $this->end->copy()->endOfDay();
$days = 0;
while($start->lt($end)) {
if (!$this->isHoliday($start) && !$this->isWeekend($start)) {
$days++;
}
$start->addDay(1);
}
return $days * $hoursInDay;
}
protected function isHoliday(Carbon $date)
{
$date->startOfDay();
foreach($this->holidays as $holiday) {
$holiday = Carbon::createFromFormat("Y-m-d", $holiday)->startOfDay();
if ($date->eq($holiday)) {
return true;
}
}
return false;
}
protected function isWeekend(Carbon $date)
{
return $date->isWeekend();
}
}
Hope this helps!
Financial years starts from April. So for example:
$startDate='2014-11-25' // Start date
$endDate ='2015-05-29' // End date
Output shows only FY-14-15 , But i want FY-14-15,FY-15-16
function calcFY($startDate,$endDate) {
$prefix = 'FY-';
$ts1 = strtotime($startDate);
$ts2 = strtotime($endDate);
$year1 = date('Y', $ts1);
$year2 = date('Y', $ts2);
$month1 = date('m', $ts1);
$month2 = date('m', $ts2);
//get months
$diff = (($year2 - $year1) * 12) + ($month2 - $month1);
/**
* if end month is greater than april, consider the next FY
* else dont consider the next FY
*/
$total_years = ($month2 > 4)?ceil($diff/12):floor($diff/12);
$fy = array();
while($total_years >= 0) {
$prevyear = $year1 - 1;
//We dont need 20 of 20** (like 2014)
$fy[] = $prefix.substr($prevyear,-2).'-'.substr($year1,-2);
$year1 += 1;
$total_years--;
}
/**
* If start month is greater than or equal to april,
* remove the first element
*/
if($month1 >= 4) {
unset($fy[0]);
}
/* Concatenate the array with ',' */
return implode(',',$fy);
}
echo calcFY('2014-11-25','2015-05-29');
My problem is, Missing Fiscal year FY-15-16. Also what i have tried is not a better code to get this for more number of years say startdate ='2014-11-25' and endDate ='2015-05-29',
$total_years returns 1 instead of 2 - you have to adjust your calculation of $diff.
function calcFY($startDate, $endDate) {
$prefix = 'FY-';
$ts1 = strtotime($startDate);
$ts2 = strtotime($endDate);
$year1 = date('Y', $ts1);
$year2 = date('Y', $ts2);
$month1 = date('m', $ts1);
$month2 = date('m', $ts2);
$fy = array();
$idx = 0;
if((int)$month1<4)
{
$fy[$idx] = $prefix . substr($year1-1, -2). '-'. substr($year1, -2);
$idx++;
}
for($i = $year1; $i<=$year2; $i++)
{
$fy[$idx] = $prefix . substr($i, -2).'-'.substr($i+1, -2); $idx++;
}
return implode(',',$fy);
}