php日期与日间之差函数-PHP源码

求二两日期之差函数

$time1 = "2008-6-15 11:49:59";//第一个时间
$time2 = "2007-5-5 12:53:28";//第二个时间
$t1 = strtotime($time1);
$t2 = strtotime($time2);
$t12 = abs($t1-$t2);
$start = 0;
$string = "两个时间相差：";
$y = floor($t12/(3600*24*360));
if($start || $y )
{
$start = 1;
$t12 -= $y*3600*24*360;
$string .= $y."年";
}
$m = floor($t12/(3600*24*31));
if($start || $m)
{
$start = 1;
$t12 -= $m*3600*24*31;
$string .= $m."月";
}
$d = floor($t12/(3600*24));
if($start || $d)
{
$start = 1;
$t12 -= $d*3600*24;
$string .= $d."天";
}
$h = floor($t12/(3600));
if($start || $h)
{
$start = 1;
$t12 -= $h*3600;
$string .= $h."时";
}
$s = floor($t12/(60));
if($start || $s)
{
$start = 1;
$t12 -= $s*60;
$string .= $s."分";
}
$string .= "{$t12}秒";
echo $string;
?>

这是一个求任意时间之差的函数

#作者：仙乐
#功能：获得任意时间与当前时间的时间差
function QueryDays($datestr){
#格式化时间
$da=preg_split("/(-| |:)/i",$datestr);
$nowyear=date("Y");
$nowmon=date("n");
$nowday=date("d");
$nowtimes=mktime(0,0,0,$nowmon,$nowday,$nowyear);
$pdtimes= mktime(0,0,0,$nowmon,$nowday,$nowyear-1);
$bjtimes= mktime(0,0,0,$da[1],$da[2],$da[0]);
#判断所给出的时间是不是在一年内
if ($bjtimes>=$pdtimes and $bjtimes<=$nowtimes){
return (floor(strftime("%j",mktime(0,0,0,$nowmon,$nowday,$nowyear)-mktime($da[3],$da[4],$da[5],$da[1],$da[2],$da[0]))));
}else{
$loop=$nowyear-$da[0];
$totaldays=(floor(strftime("%j",mktime(0,0,0,$nowmon,$nowday,$nowyear)-mktime(0,0,0,1,1,$nowyear))));
for($i=1;$i<=$loop;$i++){
for($j=12;$j>=1;$j--){
if ($da[0]==$nowyear-$i and $da[1]==$j){
$days=MonDays($nowyear-$i,$j);
return $totaldays+=$days-$da[2];
break;
}else{
$days=MonDays($nowyear-$i,$j);
$totaldays+=$days;
}//end else
}//end for
}//end for
}//end else
}//end function
#取得月分的天数
function MonDays($year,$month){
switch ($month){
case "1":
case "3":
case "5":
case "7":
case "8":
case "10":
case "12": $days=31;break;
case "4":
case "6":
case "9":
case "11": $days=30;break;
case "2":
if (checkdate($month,29,$year)){
$days=29;
}else{
$days=28;
}//end else
break;
}//end switch
return $days;
}//end function
$datestr="2002-1-14 9:47:20";
echo QueryDays($datestr);
?>