RE: [PHP-DB] Number of working hours in a month.

Hi there, Thanks, the article was posted by me at that site. I have a few different email addresses to a point I forgot which and which.. I have solved the problem. Here are the codes. <? // // 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 $weeks>5 // // Initial formula doesn't work well, so I "reversed-engineered" to get the formula. // 0 - Sunday,...,6 - Saturday for ($year = 2001; $year <= 2037; $year++) { for ($month = 1; $month <= 12; $month++) { $num_of_days = date("t", mktime(0,0,0,$month,1,$year)); echo "<HR> Month=$month Year=$year <BR>"; echo "Number of days = $num_of_days <BR>"; $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 = $firstday,$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 - $day_of_week)) + 1;
    } elseif ($firstday < $day_of_week) {
        // correct week, now move forward to specified day
        $d = ($day_of_week - $firstday + 1);
    } else {
        // my "reversed-engineered" formula
        if ($lastday==28) // max of 4 occurences each in the month of February with 28 days
           $d = ($firstday + 4);
        elseif ($firstday==4)
           $d = ($firstday - 2);
        elseif ($firstday==5 )
           $d = ($firstday - 3);
        elseif ($firstday==6)
           $d = ($firstday - 4);
        else
           $d = ($firstday - 1);
        if ($lastday==29) // only 1 set of 5 occurences each in the month of February with 29 days
           $d -= 1;
    }

    $d += 28;    // jump to the 5th week and see if the day exists
    if ($d > $lastday) {
        $weeks = 4;
    } else {
        $weeks = 5;
    }

if ($day_of_week==0) echo "Sun "; elseif ($day_of_week==1) echo "Mon "; elseif ($day_of_week==2) echo "Tue "; elseif ($day_of_week==3) echo "Wed "; elseif ($day_of_week==4) echo "Thu "; elseif ($day_of_week==5) echo "Fri "; else echo "Sat "; echo "occurences = $weeks <BR> "; } // for $day_of_week loop } // for $mth loop } // for $year loop ?> At 04:16 PM 02-03-2002 -0800, Shrock, Court wrote:

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: php-db@lists.php.net 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