I need to calculate end-date/SLA for a given start-date considering the completion-time in minutes within the business hours/days.
For instance :
start-date = 2019-03-29 15:00:00
completion-time = 720 (minutes)
Considering 8 working hours per day(9am to 5pm) and weekend on Sunday. The end-date should be 2019-04-01 11:00:00.
So the total hours are 12,
on 2019-03-29 from 15:00:00 to 17:00:00 (2 hours)
on 2019-03-30 from 09:00:00 to 17:00:00 (8 hours)
on 2019-04-01 from 09:00:00 to 11:00:00 (2 hours)
Any help with regards to this would be much appreciated.
So far i can able to get the total working days excluding the weekends but getting exact end time is where i'm struggling.
Thank you.
Update:
$bookingDateTime = Carbon::parse('2019-03-29 15:00:00');
$i = 0;
$completion_in_days = (720/60)/8; //converted minutes in no of days
$working_days = ['monday', 'tuesday', 'wednesday', 'thursday', 'friday', 'saturday'];
while($i < ($completion_in_days)){
if(array_search(strtolower($bookingDateTime->englishDayOfWeek), $working_days) !== false){
$i++;
}
$bookingDateTime->addDay();
}
This will give me the end date.
Please find the snippet below, I wrote line by line commenting for understanding
$endDate = $startDate = '2019-03-29 15:00:00';
$officeStart = '09:00:00';
$officeEnd = '17:00:00';
$totalHours = 12;
$i = 1;
$flag = false;
while ($totalHours > 0) {
$day = date('D', strtotime($endDate)); // fetching day of week
if ($day == 'Sun') { // checking if sunday thenskip by adding 1 day to end date
$endDate = date('Y-m-d', strtotime($endDate . " +1 Day")) . ' ' . $officeStart; // adding one day if sunday
continue;
}
$diff = strtotime($officeEnd) - strtotime(date("H:i:s", strtotime($endDate))); // getting difference of time of office end date and result end date
$hours = $diff / (3600); // difference in minutes
if ($hours > $totalHours) {
$hours = $totalHours;
$flag = true; // to break loop if last loop comes
} else {
$totalHours = $totalHours - $hours; // substracting hours from total hours left
}
$endDate = date('Y-m-d H:i:s', strtotime("+$hours Hour", strtotime($endDate))); // adding hours which are substracted
if (!$flag) {
$endDate = date('Y-m-d', strtotime($endDate . " +1 Day")) . ' ' . $officeStart; // if not last loop add day to result end date
} else {
break;
}
}
Output
2019-04-01 11:00:00
Demo.
Related
I like to iterate through all weeks in a date range that spawns 2 years. Starting at the current week number two years ago, until the current week number this year.
The problem is that a year can either have 52 weeks or 53 weeks, for example:
2015 had 53 weeks (the 53th was from 2015-12-28 till 2016-01-03)
2016 had 52 weeks (the 52th was from 2016-12-26 till 2017-01-01)
So this is currently my php code:
# start with the current week 2 years ago
$year = date("Y") - 2;
$week = (int) date("W"); // (int) removes leading zero for weeks < 10
$endYear = date("Y");
$endWeek = (int) date("W");
# iterate through all weeks until now
do {
echo $week. " of ". $year;
$week++;
if ($week > 52) { // or greater 53 ?????????????
$year ++;
$week = 1;
}
}
while ($year < $endYear || $week < $endWeek);
Instead of trying to keep track of the bounds, let PHP do it for you.
$start = new DateTime('-2 years');
$end = new DateTime();
$interval = new DateInterval('P1W');
$period = new DatePeriod($start, $interval, $end);
foreach ($period as $date) {
echo $date->format('W') . " of ". $date->format('Y') . "\n";
}
Demo
With the help of Ross Wilson and John Conde I found a solution:
$now = new DateTime;
$from = (new DateTime)->setISODate($now->format('Y') - 2, $now->format('W'));
$from->modify("thursday this week"); // see ISO 8601
while($from < $now) {
echo $from->format('W Y').'<br />';
$from->modify("+1 week");
}
It is important so set the week day to thursday, because according to ISO 8601, the week "belongs" to the year which still contains the thursday.
You could use DateTime and setISODate for something like this:
$now = new DateTime;
$from = (new DateTime)->setISODate($now->format('Y') - 2, $now->format('W'));
while($from < $now) {
echo $from->format('W Y').'<br />';
$from->modify("1 week");
}
Hope this helps!
I am trying to get the start date and end dates of all weeks between two week numbers.
That is one of my date is 2014-05-17 and its week number is 20 and other date is 2014-08-13 and its week number is 33.
My task is to get start and end dates of all weeks between 20 and 33. Here Sunday is the week start and Saturday week end.
$signupweek='2014-05-17';
$signupweek=date("W",strtotime($signupdate));
//week number of current date.
$weekNumber = date("W");
Can anyone help to find the dates.
try this
$signupdate='2014-05-17';
$signupweek=date("W",strtotime($signupdate));
$year=date("Y",strtotime($signupdate));
$currentweek = date("W");
for($i=$signupweek;$i<=$currentweek;$i++) {
$result=getWeek($i,$year);
echo "Week:".$i." Start date:".$result['start']." End date:".$result['end']."<br>";
}
function getWeek($week, $year) {
$dto = new DateTime();
$result['start'] = $dto->setISODate($year, $week, 0)->format('Y-m-d');
$result['end'] = $dto->setISODate($year, $week, 6)->format('Y-m-d');
return $result;
}
Output
Week:20 Start date:2014-05-11 End date:2014-05-17
Week:21 Start date:2014-05-18 End date:2014-05-24
Week:22 Start date:2014-05-25 End date:2014-05-31
Week:23 Start date:2014-06-01 End date:2014-06-07
Week:24 Start date:2014-06-08 End date:2014-06-14
Week:25 Start date:2014-06-15 End date:2014-06-21
Week:26 Start date:2014-06-22 End date:2014-06-28
Week:27 Start date:2014-06-29 End date:2014-07-05
Week:28 Start date:2014-07-06 End date:2014-07-12
Week:29 Start date:2014-07-13 End date:2014-07-19
Week:30 Start date:2014-07-20 End date:2014-07-26
Week:31 Start date:2014-07-27 End date:2014-08-02
Week:32 Start date:2014-08-03 End date:2014-08-09
Week:33 Start date:2014-08-10 End date:2014-08-16
Another method...
If you have a date, from that date you can find the start date and end date of that week. But here week number is not used.
For example:
You have a date 2014-08-13, then required start date is 2014-08-10 and end date is 2014-08-16.
PHP code is
$signupweek='2014-8-13';
/*start day*/
for($i = 0; $i <7 ; $i++)
{
$date = date('Y-m-d', strtotime("-".$i."days", strtotime($signupweek)));
$dayName = date('D', strtotime($date));
if($dayName == "Sun")
{
echo "start day is ". $date."<br>";
}
}
/*end day*/
for($i = 0; $i <7 ; $i++)
{
$date = date('Y-m-d', strtotime("+".$i."days", strtotime($signupweek)));
$dayName = date('D', strtotime($date));
if($dayName == "Sat")
{
echo "end day is ". $date."<br>";
}
}
OUTPUT
start day is 2014-08-10
end day is 2014-08-16
Hope this is useful..
Here's an example that implements the function from this answer:
$signupweek = '2014-05-17';
$signupweek = date("W", strtotime($signupweek));
$current_week = date('W');
$output = array();
// Loop through the weeks BETWEEN your given weeks
// to include the start and end week, remove the +1 below and make
// it $i <= $current_week
for($i = $signupweek + 1; $i < $current_week; $i++) {
// Get the start and end for the current week ($i)
$dates = getStartAndEndDate($i, '2014');
// if the start or end of the week is greater than now, skip it
if(strtotime($dates['start']) > time() or strtotime($dates['end']) > time())
continue;
// Add to output array
$output[] = $dates;
}
function getStartAndEndDate($week, $year)
{
$time = strtotime("1 January $year", time());
$day = date('w', $time);
$time += ((7 * $week) + 1 - $day) * 24 * 3600;
$return['start'] = date('Y-n-j', $time);
$time += 6 * 24 * 3600;
$return['end'] = date('Y-n-j', $time);
return $return;
}
Output
Try this:
$startTime = "2014-05-17";
$startWeek = 20;
$endWeek = 33;
for ($i = 0; $i <= ($endWeek - $startWeek); $i++) {
$days = $i * 7;
echo date("Y-m-d", strtotime($startTime . "+$days day")).'<br />';
}
Unfortunately, it seems that 2014-08-13 is not the start of week 33. 2014-08-16 is.
You can now use DateTime to get start/end dates of week(s)
function getDateRangeForAllWeeks($start, $end){
$fweek = getDateRangeForWeek($start);
$lweek = getDateRangeForWeek($end);
$week_dates = [];
while($fweek['sunday']!=$lweek['sunday']){
$week_dates [] = $fweek;
$date = new DateTime($fweek['sunday']);
$date->modify('next day');
$fweek = getDateRangeForWeek($date->format("Y-m-d"));
}
$week_dates [] = $lweek;
return $week_dates;
}
function getDateRangeForWeek($date){
$dateTime = new DateTime($date);
$monday = clone $dateTime->modify(('Sunday' == $dateTime->format('l')) ? 'Monday last week' : 'Monday this week');
$sunday = clone $dateTime->modify('Sunday this week');
return ['monday'=>$monday->format("Y-m-d"), 'sunday'=>$sunday->format("Y-m-d")];
}
print_r( getDateRangeForWeek("2016-05-07") );
print_r( getDateRangeForAllWeeks("2015-11-07", "2016-02-15") );
Can some one help me write a function that calculates the number of working hours between two dates but want to exclude the time when the request had a status of "On Hold".
So lets say the request came in at 3PM friday and was closed at 3PM Wednesday, and working hours are from 8AM to 5PM pacific (Mon thru Friday)...Total working hours will be 27 hours...but if the request remained on hold from Monday 3PM till Tuesday 3PM...Actual work time on the request really becomes 18 hours instead of 27 hours.
I have recently started working on PHP and have been assigned this task which is very confusing to me. Please help
All you have to do is get the total time elapsed, then substract the non-working hours.
You can use dateTime and datePeriod php objects for that (requires php 5.3)
Here a small script to do what you want (but you will have probably to adapt for your needs)
<?php
ini_set('display_errors', 'on');
define('DAY_WORK', 32400); // 9 * 60 * 60
define('HOUR_START_DAY', '08:00:00');
define('HOUR_END_DAY', '17:00:00');
// get begin and end dates of the full period
$date_begin = '2013-11-29 15:00:00';
$date_end = '2013-12-03 15:00:00';
// keep the initial dates for later use
$d1 = new DateTime($date_begin);
$d2 = new DateTime($date_end);
// and get the datePeriod from the 1st to the last day
$period_start = new DateTime($d1->format('Y-m-d 00:00:00'));
$period_end = new DateTime($d2->format('Y-m-d 23:59:59'));
$interval = new DateInterval('P1D');
//$interval = new DateInterval('weekdays'); // 1 day interval to get all days between the period
$period = new DatePeriod($period_start, $interval, $period_end);
$worked_time = 0;
$nb = 0;
// for every worked day, add the hours you want
foreach($period as $date){
$week_day = $date->format('w'); // 0 (for Sunday) through 6 (for Saturday)
if (!in_array($week_day,array(0, 6)))
{
// if this is the first day or the last dy, you have to count only the worked hours
if ($date->format('Y-m-d') == $d1->format('Y-m-d'))
{
$end_of_day_format = $date->format('Y-m-d '.HOUR_END_DAY);
$d1_format = $d1->format('Y-m-d H:i:s');
$end_of_day = new DateTime($end_of_day_format);
$diff = $end_of_day->diff($d1)->format("%H:%I:%S");
$diff = split(':', $diff);
$diff = $diff[0]*3600 + $diff[1]*60 + $diff[0];
$worked_time += $diff;
}
else if ($date->format('Y-m-d') == $d2->format('Y-m-d'))
{
$start_of_day = new DateTime($date->format('Y-m-d '.HOUR_START_DAY));
$d2_format = $d2->format('Y-m-d H:i:s');
$end_of_day = new DateTime($end_of_day_format);
$diff = $start_of_day->diff($d2)->format('%H:%I:%S');
$diff = split(':', $diff);
$diff = $diff[0]*3600 + $diff[1]*60 + $diff[0];
$worked_time += $diff;
}
else
{
// otherwise, just count the full day of work
$worked_time += DAY_WORK;
}
}
if ($nb> 10)
die("die ".$nb);
}
echo sprintf('Works from %s to %s, You worked %d hour(s)', $date_begin, $date_end, $worked_time/60/60);
Calculate work time with an accuracy of 1 minute.
WARNING: This function can take many seconds to load as it does a loop for every minute between the time span.
<?php
$request = array(
'start' => '3PM Nov 29 2013',
'end' => '3PM Dec 4 2013'
);
echo calculate_work($request);
/**
* Calculate work time by looping through every minute
* #param array $request start to end time
* #return int work time in minutes
*/
function calculate_work($request)
{
$start = strtotime($request['start']);
$end = strtotime($request['end']);
$work_time = 0;
/* Add 1 minute to the start so that we don't count 0 as a minute */
for ($time = $start + 60; $time <= $end; $time += 60)
{
// Weekends
if (date('D', $time) == 'Sat' OR date('D', $time) == 'Sun')
continue;
// Non Working Hours
if (date('Hi', $time) <= '0800' OR date('Hi', $time) > '1700')
continue;
// On Hold
if ($time > strtotime('3PM Dec 2 2013') AND $time <= strtotime('3PM Dec 3 2013'))
continue;
$work_time++;
}
// Divide by 60 to turn minutes into hours
return $work_time / 60;
}
/**
* Get the total working hours in seconds between 2 dates..
* #param DateTime $start Start Date and Time
* #param DateTime $end Finish Date and Time
* #param array $working_hours office hours for each weekday (0 Monday, 6 Sunday), Each day must be an array containing a start/finish time in seconds since midnight.
* #return integer
* #link https://github.com/RCrowt/working-hours-calculator
*/
function getWorkingHoursInSeconds(DateTime $start, DateTime $end, array $working_hours)
{
$seconds = 0; // Total working seconds
// Calculate the Start Date (Midnight) and Time (Seconds into day) as Integers.
$start_date = clone $start;
$start_date = $start_date->setTime(0, 0, 0)->getTimestamp();
$start_time = $start->getTimestamp() - $start_date;
// Calculate the Finish Date (Midnight) and Time (Seconds into day) as Integers.
$end_date = clone $end;
$end_date = $end_date->setTime(0, 0, 0)->getTimestamp();
$end_time = $end->getTimestamp() - $end_date;
// For each Day
for ($today = $start_date; $today <= $end_date; $today += 86400) {
// Get the current Weekday.
$today_weekday = date('w', $today);
// Skip to next day if no hours set for weekday.
if (!isset($working_hours[$today_weekday][0]) || !isset($working_hours[$today_weekday][1])) continue;
// Set the office hours start/finish.
$today_start = $working_hours[$today_weekday][0];
$today_end = $working_hours[$today_weekday][1];
// Adjust Start/Finish times on Start/Finish Day.
if ($today === $start_date) $today_start = min($today_end, max($today_start, $start_time));
if ($today === $end_date) $today_end = max($today_start, min($today_end, $end_time));
// Add to total seconds.
$seconds += $today_end - $today_start;
}
return gmdate("H:i:s", $seconds);
}
I need to extract the total hours in a any month, given just the MONTH and the YEAR, taking into account leap years.
Here is my code so far...
$MonthName = "January";
$Year = "2013";
$TimestampofMonth = strtotime("$MonthName $Year");
$TotalMinutesinMonth = $TimestampofMonth / 60 // to convert to minutes
$TotalHoursinMonth = $TotalMinutesinMonth / 60 // to convert to hours
Just work out the number of days in the month and then multiply by 24, like so:
// Set the date in any format
$date = '01/01/2013';
// another possible format etc...
$date = 'January 1st, 2013';
// Get the number of days in the month
$days = date('t', strtotime($date));
// Write out the days
echo $days;
You can do this:
<?php
$MonthName = "January";
$Year = "2013";
$days = date("t", strtotime("$MonthName 1st, $Year"));
echo $days * 24;
You can use DateTime::createFromFormat since you don't have day
$date = DateTime::createFromFormat("F Y", "January 2013");
printf("%s hr(s)",$date->format("t") * 24);
Well if you are looking at working day its a different approach
$date = "January 2013"; // You only know Month and year
$workHours = 10; // 10hurs a day
$start = DateTime::createFromFormat("F Y d", "$date 1"); // added first
printf("%s hr(s)", $start->format("t") * 24);
// if you are only looking at working days
$end = clone $start;
$end->modify(sprintf("+%d day", $start->format("t") - 1));
$interval = new DateInterval("P1D"); // Interval
var_dump($start, $end);
$hr = 0;
foreach(new DatePeriod($start, $interval, $end) as $day) {
// Exclude sarturday & Sunday
if ($day->format('N') < 6) {
$hr += $workHours; // add working hours
}
}
printf("%s hr(s)", $hr);
<?php
function get_days_in_month($month, $year)
{
return $month == 2 ? ($year % 4 ? 28 : ($year % 100 ? 29 : ($year %400 ? 28 : 29))) : (($month - 1) % 7 % 2 ? 30 : 31);
}
$month = 4;
$year = 2013;
$total_hours = 24 * get_days_in_month($month, $year);
?>
you can use above function to retrieve total days in a month taking into account leap year and then multiply the value to 24
plus, you can also use a cal_days_in_month function but it only supports PHP builds of PHP 4.0.7 and higher.
and if you are using the above "get_day_in_month" then you need to parse the string into integer which can be done like this
FOR MONTH
<?php
$date = date_parse('July');
$month_int = $date['month'];
?>
FOR YEAR
<?php
$year_string = "2013"
$year_int = (int) $year_string
?>
Starting with a date in this format: 2011-05-01 09:00:00, how can I create an array that contains all office hours (09:00 to 17:00) for all working days of the year (so excluding all Saturday and Sundays). What I want to arrive to is something like this:
2011-05-01 09:00:00
2011-05-01 10:00:00
2011-05-01 11:00:00
2011-05-01 12:00:00
2011-05-01 13:00:00
2011-05-01 14:00:00
2011-05-01 15:00:00
2011-05-01 16:00:00
2011-05-01 17:00:00
//next day, starting at 09:00 and ending at 17:00
2011-05-02 09:00:00
...
2011-05-02 17:00:00
//until the last day of the year from 09:00 to 17:00
2011-12-31 09:00:00
...
2011-12-31 17:00:00
The start date will be the first of the current month at with 09:00 as time and the very last date (last element of the array) will always be 17:00 on the last day of the year.
Again, weekends should be excluded.
Pseudocode idea:
I thought of something like strtotime($start, "+1 one hour") with a check for "if smaller than 17:00" but it doesn't seem to be that simple.
How about this:
$start = strtotime('2011-05-01');
$end = strtotime('2011-12-31');
$times = array();
for ($i = $start; $i <= $end; $i += 24 * 3600)
{
if (date("D", $i) == "Sun" || date("D", $i) == "Sat")
{
continue;
}
for ($j = 9; $j <= 17; $j++)
{
$times []= date("Y-m-d $j:00:00", $i);
}
}
The outer loop iterates through all the days in the given time period. In the outer loop, we check to see if the day is either Saturday or Sunday (a weekend), and if it is, we skip that day. If it's not a weekend, we loop through all the valid hours, adding the full date and time to the array as we go.
Some tips:
date("G", $some_timestamp) gives you the hour of the day in 24-hour format
date("N", $some_timestamp) gives you the day of the week, 1 (for Monday) through 7 (for Sunday)
Take a look at the php manual for date.
Edit: You can pick any start timestamp and add 3600 to add one hour, if your hour is greater than 17, you can add a bigger step to go right to the next morning, same for a weekend, and just do a while ($timestamp < $end_timestamp) {}
I'd encourage you to use the wonderful DateTime class and its related classes. Here, you can make good use of DatePeriod:
<?php
$now = new DateTime('today'); // starting time 0.00 this morning
$endOfYear = new DateTime('31 December this year 23:00'); // end time
$interval = new DateInterval('PT1H'); // frequency -- every hour
$times = array();
foreach (new DatePeriod($now, $interval, $endOfYear ) as $datetime) {
// $datetime is a DateTime object for the hour and time in question
$dow = $datetime->format('w'); // 0 is Sunday
if (($dow == '0') || ($dow == '6')) {
continue; // miss Saturday and Sunday out
}
$time = $datetime->format('G'); // hour without leading 0
if (($time < '9') || ($time > '17')) {
continue;
}
$times[] = $datetime->format('r'); // output format
}
var_dump($times);
Obviously there are various aspects of this that you can configure, especially the output format. Depending on your purpose, you may prefer to put the DateTime objects themselves into the array.
$datetime = new DateTime(); // Set your start date here
do { // Iterate over .. dont know, a long time?
do { // Iterate over the week ...
$datetime->setTime(9,0,0);
do { // Iterate over the hours ...
echo $datetime->format('c') . PHP_EOL; // Do something with $datetime here
$datetime->add(new DateInterval('PT1H'));
} while ($datetime->format('G') < 18); // .. till < 18
$datetime->add(new DateInterval('P1D')); // next day
} while (!in_array($datetime->format('w'), array('0','6'))); // until we hit sunday or saturday
$datetime->add(new DateInterval('P2D')); // next monday
} while (true); // replace with your "end" expression
Currently untested.
You can use the common interval-strings (like 1 hour and so on) too http://php.net/dateinterval.createfromdatestring
You could calculate your dates with two nested loops and generate the string with date().
I would just loop through all dates, incremented by hour, from now until the end of the year, as follows (pseudocode, obviously):
for n = now until end of year
if (date(n) is between 9:00 and 17:00) AND (if date(n) is not sat or sun)
add to array
end if
increment n by 1 hour
end
Here is a solution which should be reasonably fast since it uses no string comparisons and has only two function calls inside the loops:
function hours()
{
$start = mktime(0, 0, 0, date('n'), 1, date('Y'));
$end = mktime(0, 0, 0, 1, 1, date('Y') + 1);
$wday = date('w', $start);
$result = array();
for ($t = $start; $t < $end; $t += 3600 * 24) {
if (($wday > 0) && ($wday < 6)) {
for ($hour = 9; $hour <= 17; $hour++) {
$result[] = date('Y-m-d', $t) . sprintf(' %02d:00:00', $hour);
}
}
$wday = ($wday + 1) % 7;
}
return $result;
}