Php - 从几天到几周/天

Question

I have a number of days to a date in the future but would like to know how many weeks and days it is. Also, noting that if its less than a week, then it simply returns the same number.

Is this possible?

eg 17 days would be 2 weeks and 3 days

eg 4 days would be 4 days

Answer 1

I would try something like this:

$days = 17;
$weeks = floor($days / 7);
$dayRemainder = $days % 7;
echo $days.'<br/>'.$weeks.'<br/>'.$dayRemainder;//add whatever logic you need here to get the display the way you want it.

Answer 2

$weeks = intval($days / 7);
$days = $days % 7;

if($weeks)
{
    printf("%d weeks", $weeks);
}
if($days)
{
    if($weeks)
    {
        printf(" and ");
    }
    printf("%d days", $days);
}

Answer 3

Something along the lines of this should work

function getnumweeks(d) {
   totalDays = d;
   numWeeks = floor(d/7);
   if numWeeks != 0 {
      extraDays = totalDays % 7;
      return array(extraDays, numWeeks);
   } else {
      return array(totalDays, 0)
   }
}

Then you can call and use it as such:

ans = getnumweeks(17)

ans[0] <- Contains number of days
ans[1] <- Contains Number of Weeks

Answer 4

As the Piskvor mentioned, you should use the modulo operator:

$weeks = $days/7;
$daysleft = $days%7;

Answer 5

Let's say x is number of days, W is output value of weeks and D is output value of days remaining.

First do integer division

W = x / 7;

Then you take remainder: D = x % 7;

Answer 6

$num_days = $databack30[days_to_next_event]; 
$weeks = floor($num_days/7); 
$days = $num_days % 7;

if($weeks>'0'){ $whenitis = ' in '.$weeks.' weeks and '.$days.' days'; }

else { $whenitis = ' in '.$days.' days'; }

Answer 7

I would suggest you to re-use this powerful function datediff:

http://www.addedbytes.com/lab/php-datediff-function/

as suggested in php weeks between 2 dates

or taking inspiration from that code.