Date problem

[kevin99]

Is there any way of changing the following code so that it will not display the 31st day on months that only have 30 days.

Many thanks

$current_date = getdate();
$myhour = $current_date["hours"];
$myday = $current_date["mday"];
$mymonth = $current_date["mon"];
$myyear = $current_date["year"];

// THE PICKUP DAY
for ($i = 1; $i <= 31; $i++)
{
if ($i == $myday)
{
$pickupday[] = "<option selected>$i</option>";
}else{
$pickupday[] = "<option>$i</option>";
}
}

🙂

[lazzerous]

function daysinmonth($month, $year)
{
   $days=31;
   while(!checkdate($month, $days, $year)) $days--;
   return $days;
}

$current_date = getdate();
$myhour = $current_date["hours"];
$myday = $current_date["mday"];
$mymonth = $current_date["mon"];
$myyear = $current_date["year"];

$days_this_month = daysinmonth($mymonth, $myyear);
// ** THE PICKUP DAY **
for ($i = 1; $i <= $days_this_month; $i++)
  {
  if ($i == $myday)
    {
    $pickupday[] = "<option selected>$i</option>";
    }
  else
    {
    $pickupday[] = "<option>$i</option>";
    }
  } 

[drawmack]

The date function in php is very powerful especially when mixed with the strtotime function:
http://www.php.net/date
http://www.php.net/strtotime

$myhour = date('G');
$myday = date('j');
$mymonth = date('n');
$myyear = date('Y');
$days = date('t');


//  ** THE PICKUP DAY **
//this way is a bit more code but it's faster cause of
//no if inside the loop
for ($i = 1; $i < $days; $i++)
{
    $pickupday[] = "<option>$i</option>";
} //end for

$pickupday[] = "<option selected>$myday</option>";

for ($i = $myday+1); $i <= $days; $i++)
{
    $pickupday[] = "<option>$i</option>";
} //end for