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PHP 中的 计算2000年01月01日起到指定日的天数
出处:互联网 整理: 软晨网（RuanChen.com） 发布: 2009-09-13 浏览: 137 ::
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这是一个计算2000年01月01日起到指定日的天数的函数
<br>
(算头也算尾)日期格式为：YYYY-MM-DD
<br>
<?php
// 计算从2000年01月01日开始到某日的天数
function cal_start2end($end_day,$start_day)
{
$start_day=ereg_replace("-","",$start_day);
$end_day=ereg_replace("-","",$end_day);
if($end_day>=$start_day)
{ // 截止日期大于开始日期
if(substr($end_day,0,4)==substr($start_day,0,4))
{
if( is_int(substr($end_day,0,4)/4) )
$leap_day=29; //leap year
else
$leap_day=28;
// //同一年
if(substr($end_day,4,2)==substr($start_day,4,2))
{
// the same month
$endday_from_startday=$end_day-$start_day+1;
}
else
{
// not same month
switch (substr($end_day,4,2))
{
case "01":
$endday_from_newyear=substr($end_day,6,2)+1;
break;
case "02":
$endday_from_newyear=substr($end_day,6,2)+31;
break;
case "03":
$endday_from_newyear=substr($end_day,6,2)+$leap_day+31;
break;
case "04":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*2;
break;
case "05":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*2+30;
break;
case "06":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*3+30;
break;
case "07":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*2+30*2;
break;
case "08":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*3+30*2;
break;
case "09":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*5+30*2;
break;
case "10":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*5+30*3;
break;
case "11":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*5+30*4;
break;
case "12":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*5+30*5;
break;
}
$endday_from_startday=$endday_from_newyear;
}
return ($endday_from_startday);
}
else
{
// 不同年！
$differ_year=substr($end_day,0,4)-substr($start_day,0,4);
$how_int_4_floor=floor($differ_year/4)+1; // 向下取整
$how_int_4_ceil=ceil($differ_year/4)+1; // 向上取整
if($how_int_4_floor==$how_int_4_ceil)
$how_leap_year=$how_int_4_floor-1;
else
$how_leap_year=$how_int_4_floor;
$how_noleap_year=$differ_year-$how_leap_year;
$differ_year2day=$how_noleap_year*365+$how_leap_year*366;
if( is_int(substr($end_day,0,4)/4) )
$leap_day=29; //leap year
else
$leap_day=28;
if(substr($end_day,4,2)==substr($start_day,4,2))
{
// the same month
$endday_from_startday=$differ_year2day+ substr($end_day,4,2);
}
else
{
// not same month
switch (substr($end_day,4,2))
{
case "01":
$endday_from_newyear=substr($end_day,6,2)+1;
break;
case "02":
$endday_from_newyear=substr($end_day,6,2)+1+31;
break;
case "03":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31;
break;
case "04":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*2;
break;
case "05":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*2+30;
break;
case "06":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*3+30;
break;
case "07":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*2+30*2;
break;
case "08":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*3+30*2;
break;
case "09":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*5+30*2;
break;
case "10":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*5+30*3;
break;
case "11":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*5+30*4;
break;
case "12":
$endday_from_newyear=substr($end_day,6,2)+1+$leap_day+31*5+30*5;
break;
} //End of switch
$endday_from_startday=$endday_from_newyear+$differ_year2day;
} //End of not same month
return ($endday_from_startday);
} //End of year
} //End of $end_day>=$start_day
} //End of function
$start_day="2000-01-01";
$end_day="2011-01-01";
echo "您输入的是：".$end_day."<br>\n";
$endday_from_startday=cal_start2end($end_day,$start_day);
echo "该日距离$start_day 共 ".$endday_from_startday." 天";
?>
<br><br><br><br>时间太少了，那位大虾可以修改为计算任意两日期之间的间隔最好！！！！！
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