PHP在一周后获取日期并计算剩余天数

Question

I have a dynamic date, now what i want is that finding the date after exact one week, i have achieved that with the code below, but now i want that now many days are left for that week after date to come. i have got some sort of time stamp, but i don't know how to convert it to DAYS LEFT.

$weekDate = date( "d/m/Y", strtotime("19-05-2014") + 86400 * 7 );

echo $weekDate;// THATS PERFECT

////////////////////////////////////////////////////////////////

$future = strtotime( $weekDate ); //Future date.

$datediff = time() - $future;

$days = floor( ( ( $datediff / 24 ) / 60 ) / 60  ); //this is not perfect, returns some 

sort of timestamp

I have tried other methods which are fine, but if week completes on 26, and today is 25th it gives me 0 days left, but it should say 1 day left. please help me.

Answer 1

In your $date_diff now is less than the future date thats why its zero. Inside strtotime() function, you can directly put a relative date inside. In this case, for one week you can use +1 week or +7 days. Consider this example:

$next_week = date('d/m/Y', strtotime('19-05-2014 +1 week')); // 26/05/2014
$next_week = strtotime('19-05-2014 +7 days');
$difference = $next_week - time(); // next weeks date minus todays date
$difference = date('j', $difference);
echo $difference . (($difference > 1) ? ' days ' : ' day ') . ' left';
// should output: 1 day left