Dear all,
How do I dynamically calculate how many working hours in a particular month?
Has anyone do it before?
Here is the simple formula.
Add - Number of days in a month
Deduct - Sundays
Number of Mon-Fri in a month * 8 hours
Number of Sat in a month * 4 hours
My Figure = total working hours of a month.
I do not need to include any holidays.
This data is for my monthly timesheets.
Currently I am hard coding it while I find a solution...
I have almost managed to get the info that I wanted.
Below is the script and it works almost for all the months except for some...
I would say about 92%, 1 in 12 incorrect...
Can anyone help to debug and find a formula to get this working?
<?
$month=11;
$year=2002;
$num_of_days = date("t", mktime(0,0,0,$month,1,$year));
echo "Month=$month Year=$year <BR>";
echo "Number of days = $num_of_days <BR>";
// count how many weeks in the month have a specified day, such as Monday.
// we know there will be 4 or 5, so no need to check for $weeks<4 or $weeks>5
$firstdayname = date("D", mktime(0, 0, 0, $month, 1, $year));
$firstday = date("w", mktime(0, 0, 0, $month, 1, $year));
$lastday = date("t", mktime(0, 0, 0, $month, 1, $year));
echo "First day of the month = $firstdayname <BR> ";
for ($day_of_week = 0; $day_of_week <= 6; $day_of_week++)
{
if ($firstday > $day_of_week) {
// means we need to jump to the second week to find the first
$day_of_week
$d = (7 - ($firstday - $day_of_week)) + 1;
echo "d=$d ";
} elseif ($firstday < $day_of_week) {
// correct week, now move forward to specified day
$d = ($day_of_week - $firstday + 1);
echo "d=$d ";
} else { // $firstday = $day_of_week
// correct day in first week
$d = ($firstday - 1);
echo "d=$d ";
}
$d += 28; // jump to the 5th week and see if the day exists
echo "(Final D=$d > $lastday) ";
if ($d > $lastday) {
$weeks = 4;
} else {
$weeks = 5;
}
echo "$day_of_week occurences = $weeks <BR> ";
}
?>
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