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exclude public holiday

Posted: Tue Jul 08, 2003 2:54 am
by superman
I use javascript calendar as an input and post to another page to calculate the duration, is it possible to exclude the public holiday? Thank you.

Posted: Tue Jul 08, 2003 4:11 am
by m3mn0n
Sure.

Is it possible to post in the correct forum? =)

Posted: Tue Jul 08, 2003 4:53 am
by superman
i have the code that calculate the duration exclude saturday and sunday, but i dunno how to exclude public holiday. Can someboday help?

Posted: Tue Jul 08, 2003 7:04 am
by superman
This code is work, it take the output from another form as its input to calculate duration. I'm from Malaysia and I'm new in php, so can you help me?



<?php
function count_workdays($date1,$date2){
$firstdate = strtotime($date1);
$lastdate = strtotime($date2);
$firstday = date(w,$firstdate);
$lastday = date(w,$lastdate);
$totaldays = intval(($lastdate-$firstdate)/86400)+1;

//check for one week only
if ($totaldays<=7 && $firstday<=$lastday){
$workdays = $lastday-$firstday+1;
//check for weekend
if ($firstday==0){
$workdays = $workdays-1;
}
if ($lastday==6){
$workdays = $workdays-1;
}

}else { //more than one week

//workdays of first week
if ($firstday==0){
//so we don't count weekend
$firstweek = 5;
}else {
$firstweek = 6-$firstday;
}
$totalfw = 7-$firstday;

//workdays of last week
if ($lastday==6){
//so we don't count sat, sun=0 so it won't be counted anyway
$lastweek = 5;
}else {
$lastweek = $lastday;
}
$totallw = $lastday+1;

//check for any mid-weeks
if (($totalfw+$totallw)>=$totaldays){
$midweeks = 0;
} else { //count midweeks
$midweeks = (($totaldays-$totalfw-$totallw)/7)*5;
}

//total num of workdays
$workdays = $firstweek+$midweeks+$lastweek;

}

/*
check for and subtract and holidays etc. here
...
*/

return ($workdays);
} //end funtion count_workdays()

$date1 = "$FromDate";
$date2 = "$ToDate";

echo $Duration ;
?>