PHP calculates the number of days difference between dates (the time difference between any time and the current time)

function count_days($a,$b){
 $a_dt=getdate($a);
 $b_dt=getdate($b);
 $a_new=mktime(12,0, 0,$a_dt['mon'],$a_dt['mday'],$a_dt['year']);
 $b_new=mktime(12,0,0,$b_dt['mon'],$b_dt[ 'mday'],$b_dt['year']);
 return round(abs($a_new-$b_new)/86400);
}
//How many days are there between today and October 11, 2008
$date1= strtotime(time());
$date1=strtotime('10/11/2008');
$result=count_days($date1,$date2);
echo $result;
?>

Method 2:

//How many days are there between today and September 9, 2008
$Date_1=date("Y-m-d");
$Date_2="2008-10-11";
$d1 =strtotime($Date_1);
$d2=strtotime($Date_2);
$Days=round(($d2-$d1)/3600/24);
echo "The difference between today and October 11, 2008". $Days."Days";
?>

PHP Get the time difference between any time and the current time

Code:

#Function: Get the time difference between any time and the current time
function QueryDays($datestr){
#Format time
$da=preg_split("/(-| |:)/i ",$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]);
#Judge whether the time given is within one year
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
#Get the number of days in the month
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);
?>

PHP Calculate the year, month, hour, minute and second difference between two times

How to calculate the difference between two times in years, months, hours, minutes and seconds in PHP

First convert the two times into timestamps, then subtract them to get the seconds difference between the two times, and finally do some arithmetic to get the year, month, day, hour, minute and second difference between the two times.

Code:

$time1 = "2008-6-15 11:49:59"; //The first time
$time2 = "2007-5-5 12:53:28"; //The second time
$t1 = strtotime($time1);
$t2 = strtotime($time2);
$t12 = abs($t1-$t2);
$start = 0;
$string = " The difference between the two times: ";
$y = floor($t12/(3600*24*360));
if($start || $y )
{
$start = 1;
$t12 -= $y *3600*24*360;
$string .= $y."year";
}
$m = floor($t12/(3600*24*31));
if($start || $m)
 {
$start = 1;
$t12 -= $m*3600*24*31;
$string .= $m."month";
}
$d = floor($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."hour";
 }
$s = floor($t12/(60));
if($start || $s)
{
$start = 1;
$t12 -= $s*60;
$string .= $s ."minutes";
}
$string .= "{$t12} seconds";
echo $string;
?>