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 $year loop
?>
At 04:16 PM 02-03-2002 -0800, Shrock, Court wrote:
>I just ran across this description[1] that points to this link[2]. HAven't
>tried it personally, and the site requires a login, so you might not find it
>useful, but just in case.
>
>[1]Count how many weeks in the month have a specified day, such as Mon, Tue,
>etc. Var avail - number of days - first dayname of the month, occurences of
>Sun, occurences of Mon, etc. Allows you to calculate number of working hours
>exclude Holidays.
>[2] http://www.weberdev.com/get_example.php3?count=3267
>
>-----Original Message-----
>From: Hoo Kok Mun
>To: [EMAIL PROTECTED]
>Sent: 3/2/02 1:31 AM
>Subject: [PHP-DB] Number of working hours in a month.
>
>Dear all,
>
>How do I dynamically calculate how many working hours in a particular
>month?
>Has anyone do it before?
>
>Here is the simple formula.
>Add - Number of days in a month
>Deduct - Sundays
>Number of Mon-Fri in a month * 8 hours
>Number of Sat in a month * 4 hours
>My Figure = total working hours of a month.
>I do not need to include any holidays.
>
>This data is for my monthly timesheets.
>Currently I am hard coding it while I find a solution...
>
>I have almost managed to get the info that I wanted.
>Below is the script and it works almost for all the months except for
>some...
>I would say about 92%, 1 in 12 incorrect...
>
>Can anyone help to debug and find a formula to get this working?
>
><?
>$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> ";
> }
>?>
>
>
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