你可以试试下面的代码:

$dt1 = strtotime("2019-12-12"); //Enter your first date
$dt2 = strtotime("12-12-2020"); //Enter your second date
echo abs(($dt1 - $dt2) / (60 * 60 * 24));

如果你正在使用MySql

function daysSince($date, $date2){
$q = "SELECT DATEDIFF('$date','$date2') AS days;";
$result = execQ($q);
$row = mysql_fetch_array($result,MYSQL_BOTH);
return ($row[0]);

}

function execQ($q){
$result = mysql_query( $q);
if(!$result){echo ('Database error execQ' . mysql_error());echo $q;}    
return $result;

}

你可以试试下面的代码:

$dt1 = strtotime("2019-12-12"); //Enter your first date
$dt2 = strtotime("12-12-2020"); //Enter your second date
echo abs(($dt1 - $dt2) / (60 * 60 * 24));

function howManyDays($startDate,$endDate) {

    $date1  = strtotime($startDate." 0:00:00");
    $date2  = strtotime($endDate." 23:59:59");
    $res    =  (int)(($date2-$date1)/86400);        

return $res;
} 

如果你有以秒为单位的时间(即unix时间戳)，那么你可以简单地减去时间并除以86400(秒/天)

$datediff = floor(strtotime($date1)/(60*60*24)) - floor(strtotime($date2)/(60*60*24));

如果需要的话:

$datediff=abs($datediff);