I'm trying to generate random date, any day from 10 to 30 days after the current day and the generated day should not be Saturday and Sunday. I search quite a lot but haven't found the solution for this.
Try:
repeat
day <- pickRandomDay(10 ... 30)
until (not satOrSun(day))
return day
You won't get too many times round the loop.
Related
What I have is:
$period = CarbonPeriod::create("03-09-2022", '1 month', "20-03-2023"); // 3rd of September and 20th of March
From this I can tell the number of months on which the period spans. But what I need is to have a way to calculate a salary based on that period knowing a monthly salary.
So what I'm actually trying to get is:
28 Days(for the first month) followed by 5 full months followed by 20 days (for the last month)
Is there any way I can get this from CarbonPeriod or CarbonInterval?
The following will give you a good approximation:
$salaryPerMonth = 1000;
$totalSalary = Carbon::create('03-09-2022')->floatDiffInRealMonths('20-03-2023') * $salaryPerMonth;
But unless you're working 7/7 days, the approach is not exact, you would need to take into account only business days (excluding, holidays, etc.)
If you want to go to this deeper precision level, take a look at https://github.com/kylekatarnls/business-day and https://github.com/kylekatarnls/business-time
I am trying to write a php solution to calculate the planned end time considering the target in business hours.
It shouldn't consider some days (retrieved from setting saved in db) such as holidays.
Also business hours are retrieved from db (morning_from (8:30am), morning_to (1:00pm), evening_from (2:30pm), evening_to (6:30pm)).
I want to develop this script because I want that my page shows the remaining time for technical resolution of an opened ticket every day.
For example:
customer having contract with 10 working hours SLA opens a ticket
today (friday) 31/01/2020 16:00:00, considering that in the
noBusinessDays = array("saturday", "sunday") and businessHours set as mentioned before(8:30-13:00/14:30-18:30), the result will have to
be monday 3/02/2020 17:30:00.
Code example:
$noBusinessDays = array("saturday", "sunday");
$businessHours = array("morning_from" => "8:30", "morning_to" => "13:00", "evening_from" => "14:30", "evening_to" => "18:30");
$SLA = "10"; //hours
$ticketDate = new DateTime();
$ticketDate->setTimestamp(strtotime("31/01/2020 16:00:00"));
// I don't know how to use my arrays to say in this calculation how to use them
$maximumLimit = $ticketDate->add(new DateInterval("PT" . $SLA ."H"));
Thank you in advance.
You may use the following function
// intersection of 2 time intervals
// input - Unix timestamps (start,end)
// output - intersection in seconds
function time_union($b_1,$e_1,$b_2,$e_2)
{
return max(0,$e_1-$b_1 - max(0,$e_1-$e_2) - max(0,$b_2-$b_1));
}
You will start with an empty time interval [X, Y) where X is the timestamp of the ticket creation and Y initially is equal to X.
Then you start adding days to Y - one by one. Each time you expand the time interval to contain another day - you use the above function to check how much of the SLA hours are covered (i.e. overlapping) with the working hours in the day you have just added. If the day is marked as a holiday - you simple skip it and continue with the next date.
If you find out that SLA hours are partially covered with either the morning or evening business hours - you should simply subtract the extra hours.
In the end Y will be equal to the timestamp that you want to show in your application.
I think I'd break down the problem into pieces. After calculating the total number of days in the interval, first dispose of the trivial case that it's all happening in one week.
begin by calculating the number of "whole weeks." Each "whole week" is five business days. Subtract the corresponding interval of time and proceed. Now, look at the day-of-the-week of the start-date: each day adds a certain number of days. Then the day-of-week of the end date, likewise. You can then consider hour-of-the-day as needed.
Holidays are a simple table: if the day falls within the range, subtract one day.
Now ... having said all of that, the very first thing that I would do is to search GitHub and SourceForge! Because I am extremely sure that somebody out there has already done this. :-D
"Actum Ne Agas: Do Not Do A Thing Already Done."
I have the following example of me subtracting the DateInterval from DateTimeImmutable
$dateA = new DateTimeImmutable('2016-06-30');
$dateB = new DateTimeImmutable('2016-05-31');
$dateInterval = new DateInterval('P3M');
// print 2016-03-30 as expected
echo $dateA->sub($dateInterval)->format('Y-m-d');
// print 2016-03-02 which i would expect 2016-02-29
echo $dateB->sub($dateInterval)->format('Y-m-d');
When I set the period to 'P8M' it works as expected. How it comes, it dosent works for february?
Ok, it's really simple (kind of). Each 'month' interval evaluates to the prior (or X number of prior) month's equivalent day. If there are more days in the current month, than the month being landed on, the excess overflows to the following month.
So if you have a date which is May 31, 2016 and want to subtract 3 month intervals, it will:
Go back 3 months (in the list of months, don't think days yet), resulting in 'February'
Then look for February 31st. This doesn't exist so bleed over to following month 2 days (2016 Febuary has 29 days, so 2 extra days)
Viola! March 2nd.
Go forward, lets say you're in May 31, 2016 and want to add one month
Go forward one month to June.
Look for June 31st, nope, 1 extra day, bleed over to July.
As expected, July 1st is the answer.
The lesson in this: Adding and Subtracting Month intervals sucks, is confusing, and can lead to non-intuitive results unless you've got your month calculation rosetta stone with you.
Explanation from the PHP Docs
Note:
Relative month values are calculated based on the length of months that they pass through. An example would be "+2 month 2011-11-30", which would produce "2012-01-30". This is due to November being 30 days in length, and December being 31 days in length, producing a total of 61 days.
In PHP, I'm generating a calendar with code from here. I've connected it with Basecamp and am successfully calculating the hours worked each day and sum them for the week and month. To properly calculate overtime, I need the hours from the whole week, even when it spans the end/beginning of a month.
After several hours of trying, I haven't been able to calculate the dates of the days marked with '?' in the above image.
I can't escape the feeling that I'm missing something obvious here, but haven't been able to figure it out. Any help much appreciated, thank you.
You can use mktime() and set a day argument which is less than 1: 0 for the last day, -1 for the day before the last day, etc.
That gives you a UNIX timestamp you can use to find out the weekday and then - until you've reached Sunday/Monday (whatever week start you use) - subtract -86400 or call mktime() again with the day reduced by 1.
Example:
mktime(0, 0, 0, 12, -1, 2010); # timestamp of Nov 29th 00:00:00 2010
I need to calculate a difference between a starting date/time and an ending date/time. But, I only want to do this for the 5-day work week (exclude Sat/Sun as days). What is the best way to do this? My thought is that from the date, I'll have to get the day of the week and if it is a weekday, then I add to the accumulator. If it's not, then I don't add anything.
I'm sure someone has done this before, but I couldn't seem to find anything searching. Any links or other help would be very useful.
Many thanks,
Bruce
DAYOFWEEK returns 1 for Sunday and 7 for Saturday. I'm not sure how your schema is set up, but this will perform a TIMEDIFF of two dates that are on a Monday - Friday work week.
select TIMEDIFF(date1,date2) from table
where DAYOFWEEK(date1) not in (1,7) and DAYOFWEEK(date2) not in (1,7)
MySQL DATE/TIME functions
EDIT: From Bruce's comment about holidays. If you have a table full of holiday dates, something like this would work to exclude processing those days:
select TIMEDIFF(date1,date2) from table
where date1 not in (select holiday from holiday_table) and
date2 not in (select holiday from holiday_table) and
DAYOFWEEK(date1) not in (1,7) and DAYOFWEEK(date2) not in (1,7)
NETWORKDAYS() "Returns the number of whole working days between start_date and end_date. Working days exclude weekends and any dates identified in holidays. Use NETWORKDAYS to calculate employee benefits that accrue based on the number of days worked during a specific term." according to the Excel 2007 help file.
The "between" description is a bit inaccurate because it includes the start and end dates, i.e. networkdays(21-01-2010. 22-01-2010) = 2. It also takes no account of times.
Here's a function in PHP that will give the same results. It doesn't work properly if the end date is less than the start date, nor does do anything about holidays (see below the function).
function networkdays($startdate, $enddate)
{
$start_array = getdate(strtotime($startdate));
$end_array = getdate(strtotime($enddate));
// Make appropriate Sundays
$start_sunday = mktime(0, 0, 0, $start_array[mon], $start_array[mday]+(7-$start_array[wday]),$start_array[year]);
$end_sunday = mktime(0, 0, 0, $end_array[mon], $end_array[mday]- $end_array[wday],$end_array[year]);
// Calculate days in the whole weeks
$week_diff = $end_sunday - $start_sunday;
$number_of_weeks = round($week_diff /604800); // 60 seconds * 60 minutes * 24 hours * 7 days = 1 week in seconds
$days_in_whole_weeks = $number_of_weeks * 5;
//Calculate extra days at start and end
//[wday] is 0 (Sunday) to 7 (Saturday)
$days_at_start = 6 - $start_array[wday];
$days_at_end = $end_array[wday];
$total_days = $days_in_whole_weeks + $days_at_start + $days_at_end;
return $total_days;
}
To take holidays into account, you'd have to work out the number of days using this function, then use a query like
Select count (holiday_date) from holidays
where holiday_date between start_date and end_date
and DAYOFWEEK(holiday_date) not in (1,7)
Be careful that there isn't a problem with the end_date being treated as 00:00 (i.e. first thing in the morning) - you may have to condition it to be 23:59:59 so that it works properly. It all depends on how your holidays are stored.
To return the holidays in the same time period and subtract that from the number you first thought of.