php中strtotime函数用法详解

<?php
//function
function nextWeeksDay($date_begin,$nbrweek)
{
$nextweek=array();
for($i = 1; $i <= $nbrweek; $i++)? { // 52 week in one year of coursewww.phpfensi.com
$nextweek[$i]=date('D d M Y',strtotime('+'.$i.' week',$date_begin));
}
return $nextweek;
}
/// end function?
/// example of a select date?
// var
$date_begin = strtotime('06-05-2010'); //D Day Month Year? - like function format.
$nbrweek=52;
// call function
$result=nextWeeksDay($date_begin,$nbrweek);
// Preview?
for($i = 1; $i <= $nbrweek; $i++)? {
echo '<br> - '.$result[$i];
}
?>
?
<?php
$str = 'Not Good';
?
// previous to PHP 5.1.0 you would compare with -1,instead of false
if (($timestamp = strtotime($str)) === false) {
??? echo "The string ($str) is bogus";
} else {
??? echo "$str == " . date('l dS o F Y h:i:s A',$timestamp);
}
?>?
<?php
echo strtotime("now")," ";
echo strtotime("10 September 2000")," ";
echo strtotime("+1 day")," ";
echo strtotime("+1 week")," ";
echo strtotime("+1 week 2 days 4 hours 2 seconds")," ";
echo strtotime("next Thursday")," ";
echo strtotime("last Monday")," ";
?>

function get_working_days($to_date) {
??? $holidays = array(
??????? 1 => array(10),//2011 ...
??????? 2 => array(11),
??????? 3 => array(21),//... 2011
??????? 4 => array(29,30),//2010 ...
??????? 5 => array(3,4,5),
??????? 6 => array(),
??????? 7 => array(19),
??????? 8 => array(11,12,13),
??????? 9 => array(20,23),
?????? 10 => array(11),
?????? 11 => array(3,
?????? 12 => array(23) //... 2010
??? );
?
??? for($to_date,$w = 0,$i = 0,$x = time(); $x < $to_date; $i++,$x = strtotime("+$i day")) {
?????? if(date("N",$x) < 6 &! in_array(date("j",$x),$holidays[date("n",$x)])) $w++;
??? }
??? return $w;
}
?
//Usage:
echo get_working_days(strtotime("2011-01-08"));