I just ran across this description[1] that points to this link[2]. HAven't tried it personally, and the site requires a login, so you might not find it useful, but just in case.
[1]Count how many weeks in the month have a specified day, such as Mon, Tue, etc. Var avail - number of days - first dayname of the month, occurences of Sun, occurences of Mon, etc. Allows you to calculate number of working hours exclude Holidays. [2] http://www.weberdev.com/get_example.php3?count=3267 -----Original Message----- From: Hoo Kok Mun To: [EMAIL PROTECTED] Sent: 3/2/02 1:31 AM Subject: [PHP-DB] Number of working hours in a month. 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> "; } ?> -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
