PHP Strtotime与变量，没有时间

Question

I am using this notation of strtotime without setting time which seems to work fine but I was wondering if it's valid or if there is a more appropriate syntax.

$now = time();
$date = strtotime("$sryear-$srmon-$srday ");
$datediff = $date - $now;

I am just calculating the difference between today and the set date and I would like to ignore time. $sryear, $srmon, $srday are variables containing year, month, day respectively.

Answer 1

how about this?

$now = date('Y-m-d');
$datediff = abs(strtotime($now) - strtotime($sryear."-".$srmon".-".$srday));
$days = $datediff/(60 * 60 * 24);

printf("Days difference between %s and %s = %d", $sryear."-".$srmon".-".$srday, $now, $days);