how to calculate days until next birthday using php - php

Question

I have 2 input fields for birth dates. I would like to then calculate days until the next birthday for each individual and the difference between both dates entered to determine who is older. Is my syntax logical/where did i go wrong?
<form>
John's Birthday (YYYY-MM-DD):
<input type=text name=jhbd value=0><br> Jake's Birthday (YYYY-MM-DD):
<input type=text name=jkbd value=0><br>
<input type=submit>
</form>
$john_bd = $_POST['jhbd'];
$jake_bd = $_POST['jkbd'];
$today = date("Y/m/d");
$interval_jh = $john_bd ->diff($today);
$interval_jk = $jake_bd ->diff($today);
echo "There are".$interval_jh->days."days until John's birthday ";
echo "There are".$interval_jk->days."days until Jake's birthday ";
if ($john_bd > $jake_bd) {
echo "John is older";
} else if ($jake_bd > $john_bd) {
echo "Jake is older";
} else {
echo "Both Jake and John are twins!";
}

Answer 1

You need to convert them first.
$today = time();
$your_date = strtotime($john_bd);
$datediff = $now - $john_bd;
echo floor($datediff/(60*60*24));
it might help you

Answer 2

Your variant is almost correct. Use format for show diff between dates. Like in example
$origin = new DateTime('2009-10-11');
$target = new DateTime('2009-10-13');
$interval = $origin->diff($target);
echo $interval->format('%R%a days');

Answer 3

public function days_to_birth(string $date) : int {
if(empty($date))
{
return -1;
}
if($date == '0000-00-00')
{
return -1;
}
$ts = strtotime($date.' 00:00:00');
$bY = date('Y',$ts);
$bm = date('m',$ts);
$bd = date('d',$ts);
$nowY = date('Y');
$nowm = date('m');
$nowd = date('d');
if($bm == $nowm && $bd >= $nowd)
{
return $bd - $nowd;
}
if( ($bm == $nowm && $bd < $nowd) || ($bm < $nowm) )
{
$nextBirth = ($nowY+1).'-'.$bm.'-'.$bd;
$nextBirthTs = strtotime($nextBirth);
$diff = $nextBirthTs - time();
return floor($diff/(60*60*24));
}
if($bm > $nowm )
{
$nextBirth = $nowY.'-'.$bm.'-'.$bd.'00:00:00';
$diff = strtotime($nextBirth) - time();
return floor($diff/(60*60*24));
}
return -1;
}