php 日期与日间之差函数

<script>ec(2);</script>

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 $bjtimes         return (floor(strftime("%j",mktime(0,0,0,$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;$i             for($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);
?>