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));

如果你想在开始日期和结束日期之间重复所有的日子，我想出了这个:

$startdatum = $_POST['start']; // starting date
$einddatum = $_POST['eind']; // end date

$now = strtotime($startdatum);
$your_date = strtotime($einddatum);
$datediff = $your_date - $now;
$number = floor($datediff/(60*60*24));

for($i=0;$i <= $number; $i++)
{
    echo date('d-m-Y' ,strtotime("+".$i." day"))."<br>";
}

<?php
$date1=date_create("2013-03-15");
$date2=date_create("2013-12-12");
$diff=date_diff($date1,$date2);
echo $diff->format("%R%a days");
?>

上面的代码用的很简单。谢谢。

你可以通过简单的方法找到约会对象

<?php
$start  = date_create('1988-08-10');
$end    = date_create(); // Current time and date
$diff   = date_diff( $start, $end );

echo 'The difference is ';
echo  $diff->y . ' years, ';
echo  $diff->m . ' months, ';
echo  $diff->d . ' days, ';
echo  $diff->h . ' hours, ';
echo  $diff->i . ' minutes, ';
echo  $diff->s . ' seconds';
// Output: The difference is 28 years, 5 months, 19 days, 20 hours, 34 minutes, 36 seconds

echo 'The difference in days : ' . $diff->days;
// Output: The difference in days : 10398

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;

}

这个工作!

$start = strtotime('2010-01-25');
$end = strtotime('2010-02-20');

$days_between = ceil(abs($end - $start) / 86400);