PHP Show Payroll Date - php

Question

I have a page that needs to show a rolling payroll type date. The payroll in question runs 1-15th and 16th - End of month. I have a page that shows the current payroll with a button to go forward or backward. I have the following code which works most of the time, but occasionally skips pay periods. Is there a better way to do this?
//find out what the offset is for the pay period we want to see
$offset = $_GET['offset'];
$offsetdays = $offset * 15;
//Find out what day it is
$dayofmonth = date('j', mktime( 0, 0, 0,date("m"), -$offsetdays, date("Y")));
//if it is the second half of the month set the pay period to be the 1-15th of this month
if($dayofmonth <= 15){
//Echo display dates in HTML
Echo "<div style='width:15%; float:right; text-align:center'>Pay Period End Date <BR>";
echo date("m/d/Y", mktime(0, 0, 0, date("m", mktime( 0, 0, 0,date("m"), -$offsetdays, date("2014"))), 15, date("2014"))); //End Date
Echo "</div> <div style='width:1%; float:right; text-align:center'>|<BR>|</div>";
Echo "<div style='width:15%; float:right; text-align:center'>Pay Period Start Date <BR>";
Echo date("m/d/Y", mktime(0, 0, 0, date("m", mktime( 0, 0, 0,date("m"), -$offsetdays, date("2014"))), 1, date("2014"))); //Start Date
Echo "</div>";
Echo "<BR>";
//Set variable dates for SQL
$checkoffset = offsetdays - 15;
$payrollstartdate = date("Y-m-d", mktime(0, 0, 0, date("m", mktime( 0, 0, 0,date("m"), - $offsetdays, date("2014"))), 1, date("Y")));
$payrollenddate = date("Y-m-d", mktime(0, 0, 0, date("m", mktime( 0, 0, 0,date("m"), - $offsetdays, date("2014"))), 15, date("Y")));
$checkdayte = date("m/d/Y", mktime(0, 0, 0, date("m", mktime( 0, 0, 0,date("m"), - $checkoffset, date("2014")))+1, 0, date("Y")));
}
//if it is the first half of the month set the pay period to be the 16- end of this month
if($dayofmonth > 15){
//Echo display dates in HTML
Echo "<div style='width:15%; float:right; text-align:center'>Pay Period End Date <BR>";
echo date("m/d/Y", mktime(0, 0, 0, date("m", mktime( 0, 0, 0,date("m"), -$offsetdays, date("2014")))+1, 0, date("2014"))); //End Date
Echo "</div> <div style='width:1%; float:right; text-align:center'>|<BR>|</div>";
Echo "<div style='width:15%; float:right; text-align:center'>Pay Period Start Date <BR>";
echo date("m/d/Y", mktime(0, 0, 0, date("m", mktime( 0, 0, 0,date("m"), -$offsetdays, date("2014"))), 16, date("2014"))); //Start Date
Echo "</div>";
Echo "<BR>";
//Set variable dates for SQL
$checkoffset = offsetdays - 15;
$payrollstartdate = date("Y-m-d", mktime(0, 0, 0, date("m", mktime( 0, 0, 0,date("m"), - $offsetdays, date("2014"))), 16, date("2014")));
$payrollenddate = date("Y-m-d", mktime(0, 0, 0, date("m", mktime( 0, 0, 0,date("m"), - $offsetdays, date("2014")))+1, 0, date("2014")));
$checkdayte = date("m/d/Y", mktime(0, 0, 0, date("m", mktime( 0, 0, 0,date("m"), -$checkoffset, date("2014"))), 15, date("2014")));
}
Increasing the offset by 1 will show the next pay period, while decreasing it by one will show the last.
I have also attempted to do this in MySQL without much luck.
In a little more detail here is what I would like to happen:
Today(2/27) when I go to the the page I need the start date to be 2/15/2014, and the end date to be 2/28/2014. When I increase the offset variable I need the start date to be 3/1/2014, and the end date to be 3/15/2014.
Next Tuesday(3/4) when I go to this page I need the start date to be 3/1/2014, and the end date to be 2/15/2014. When I increase the offset variable I need the start date to be 3/15/2014, and the end date to be 3/31/2014.

Answer 1

Here is a MySQL-centered way of solving this problem. It's based on the idea that date intervals are efficiently described using the starting date of each interval.
The time intervals you need for your payroll business rules are pay periods -- "half" months -- defined as starting on YYYY-mm-01 and YYYY-mm-16.
So, given an arbitrary timestamp, for example NOW() or 1941-12-07 08:00, we need a way to determine the starting day of the pay period. It would be handy to have a MySQL stored function to do that, in such a way that
BIMONTHLY_PAY_PERIOD_START('1941-12-07 08:00') produces '1941-12-01'
and
BIMONTHLY_PAY_PERIOD_START('2014-03-20 20:50') produces '2014-03-16'
Here's a definition of that stored function.
DELIMITER $$
DROP FUNCTION IF EXISTS `BIMONTHLY_PAY_PERIOD_START`$$
CREATE FUNCTION `BIMONTHLY_PAY_PERIOD_START` (datestamp DATETIME)
RETURNS DATE
NO SQL
DETERMINISTIC
COMMENT 'returns first day of bimonthly pay period: 1,16 of month'
BEGIN
DECLARE dy INT;
SET dy = DAY(datestamp);
IF dy<=15
THEN SET dy=0;
ELSE SET dy=15;
END IF;
RETURN (DATE(DATE_FORMAT(datestamp, '%Y-%m-01')) + INTERVAL dy DAY);
END$$
DELIMITER ;
It does it, basically, by finding the first day of the month and then offsetting it by 15 days if necessary.
Next, we need a function, given a pay period (remember, it's defined by its first day), to move forward or backward a particular number of pay periods. For example
BIMONTHLY_PAY_PERIOD_ADD('1941-12-01', -1) goes back one period to '1941-11-16'
and
BIMONTHLY_PAY_PERIOD_ADD('2014-03-16', 3) goes forward three to '2014-05-01'
Here is a definition of that function:
DELIMITER $$
DROP FUNCTION IF EXISTS `BIMONTHLY_PAY_PERIOD_ADD`$$
CREATE FUNCTION `BIMONTHLY_PAY_PERIOD_ADD`(datestamp DATETIME, incr INT)
RETURNS DATE
NO SQL
DETERMINISTIC
COMMENT 'returns first day of a pay period in the future or past'
BEGIN
DECLARE pp DATE;
SET pp = BIMONTHLY_PAY_PERIOD_START(datestamp);
IF ABS(incr) MOD 2 = 1
THEN
SET pp = pp + INTERVAL 16 DAY;
SET incr = incr - 1;
END IF;
SET incr = incr DIV 2;
SET pp = pp + INTERVAL incr MONTH;
RETURN BIMONTHLY_PAY_PERIOD_START(pp);
END$$
DELIMITER ;
It works stably, without skipping pay periods, by using the insight that there are two pay periods in each month.
Finally, you need a way to use these pay period dates in SQL logic. For example, suppose you have a table of payroll records containing
employee_id INT identifies the employee
clock_in DATETIME the date and time the employee clocked in
clock_out DATETIME the date and time the employee clocked out
and you want to summarize minutes-on-the-job by employee and pay period for the last six complete pay periods. That is, you want to exclude minutes-on-the-job for the present, incomplete, pay period.
You do this:
SELECT employee_id,
BIMONTHLY_PAY_PERIOD_START(clock_out) AS pay_period,
SUM(TIMESTAMPDIFF(MINUTE,clock_in, clock_out)) AS minutes_worked
FROM payroll_data
WHERE clock_out >= BIMONTHLY_PAY_PERIOD_ADD(NOW(), -6)
AND clock_out < BIMONTLY_PAY_PERIOD_START(NOW())
GROUP BY employee_id, BIMONTHLY_PAY_PERIOD_START(clock_out)
ORDER BY employee_id, BIMONTHLY_PAY_PERIOD_START(clock_out)
This query uses the functions we've defined to figure out which pay period each day's work is in. Notice that it uses the >= comparison operator for the start of the date range it needs, and the < operator combined with the day after the date range to handle the end of the date range.
If you need the last day of the present pay period, that's easy. It's the day before the start of the next pay period.
BIMONTHLY_PAY_PERIOD_ADD(NOW(),1) - INTERVAL 1 DAY
This way of handling date arithmetic is robust and reasonably fast. It exploits useful features of MySQL like date-range index searches and aggregate (GROUP BY) queries. You should be able to combine it with client code (php) to make nice presentations of your information.

Answer 2

One of the resons that could cause skipping periods is your way of computing $offsetdays. All months aren't 30 days long, so multiplying the period offset by 15 won't always work, especially with big offsets.
<?php
// Set default timezone to avoid warnings
date_default_timezone_set("UTC");
function get_period_bounds($offset = 0) {
$secondhalf = ($offset % 2) == 0 xor (int) date('j') >= 15;
$monthnumber = ceil((int) date('n') + $offset / 2);;
$period_begin = mktime(0, 0, 0, // 00:00:00
$monthnumber,
$secondhalf ? 16 : 1);
$period_end = mktime(0, 0, 0, // 00:00:00
$secondhalf ? $monthnumber + 1 : $monthnumber,
$secondhalf ? 0 : 15);
return array($period_begin, $period_end);
}
The magic
Let's explain the tiny bits of magic involved here :
Month half
To determine if the target period will be in the second half of the month, we first need to know whether today is on the first or second half of the month. That's done with (int) date('j') >= 15. We simply convert the day number to an integer and compare it to the threshold (15).
If today is on the second half, the next period will be on the first half, the one after on the second, the one after on the first, etc. So if I skip an even number of periods (i.e. a round number of months), the target period is still on the second half. This applies with a negative offset (go backward) and easily adapts if today is on the first half.
Hence the target period is the second half if :
Today is in the first half and we skip an odd number of periods ;
or today is in the second half and we skip an even number of periods.
We can combine these two conditions into one using the XOR (eXclusive OR) operator :
target_in_second_half = (today in second half) XOR (even offset)
Month
Now that we know in which half of the month the target period is, we need to know in which month is is. Since there are exactly two periods in a month, half of the offset is the month offset. So we add half the offset to the current month and ceil the result. Most of the magic happens in the mktime function that understands for instance than month #0 is the last month of the previous year.
Begin date
The beginning date is pretty straightforward : we simply use the relative month number with the current year (used by default in mktime). If the target period is on the first period, the first day of the period is the 1st, and 16th otherwise.
I used the ternary operator for readability :
(BOOLEAN TEST) ? VALUE_IF_TRUE : VALUE_IF_FALSE
End date
The case of the first half is straightforward, we use the relative month index and the end day is the 15th. For the second half, the last little trick is to get the number of the last day of the month. To do so, we rely (again) on mktime's ability to transform relative second/minute/hour/day/month/year by adding one to the month number and setting the day number to 0, i.e. the last day of the previous month. To sum up, we get the last day of the month before the month after the period month, i.e. the last day of the month of the period.
Check
You may want to check if this formula doesn't skip a period once in a while. This little loop will help you do that :
<?php
$len = 500; // Change to your needs
$last = 0;
for ($i = -$len; $i <= $len; $i++) {
$a = get_period_bounds($i);
printf("Period %d is from %s to %s [%s]<br />",
$i, // Period number
date('Y-m-d', $a[0]), // Begin date
date('Y-m-d', $a[1]), // End date
$last + 3600 * 24 == $a[0] ? 'x' : ' ' // Prints a x if the last day of the previous period is the day before the first day of this period
);
$last = $a[1];
}
This snippet will print the period dates between -$len and $len, with a checkmark [x] if the last tay of the period before is really the day before the first day of the current period, i.e. whether or not the periods really are adjacent.
(note : there won't be a check on the first line because there's no previous data to compare)