php计算日期相差天数(任意时间与当前时间的时间差)

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方法2：

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PHP获得任意时间与当前时间的时间差

代码：

#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 $bjtimesreturn (floor(strftime("%j",mktime(0,0,0,$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;$ifor($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);
?>