[SOLVED] Date() with Leap Year Question

[phpQuestioner]

I am using this function below in a script to create a count down script.

 

<?
date(("z"), time() - (14400 * 1))
?>

 

My question is; does the "z" (0 - 365) go up to a maxium of 366 when it is leap year?

[dustinnoe]

<?php
// leap year to test
$time = strtotime("Dec 31, 2004");
$julian  = date('z', $time);
echo $julian;
?>

This will output '365' but Jan 1st will output '0' so the answer to your question is yes and no at the same time.

[phpQuestioner]

ok so; let me give you an example

 

<?php
$doy = date(("z"), time() - (14400 * 1));
$calculate = 365 - $doy;
$currentyear = date(("Y"), time() - (14400 * 1));
$output = $currentyear + 1;

if ($calculate == 0)
{
echo "Happy New Year";
}
else if ($calculate <= 365)
{
echo "There are $calculate days left until $output.";
}
?>

 

Now I know I probably could set up code to check for leap year. But the question now posed, is will leap year automatically be configured into the 365 or will I have to create some code too look for leap year; that will add 365 + 1 (or something like that), if leap year occurs?

[kenrbnsn]

Just check the day number for the last day of the year. It will be 365 for leap years and 364 for non-leap years.

 

Ken

[phpQuestioner]

Thanks for the help dustinnoe and kenrbnsn; I came up with a little bit of a different solution.

 

 

<?php

$check4leap = date(("L"), time() - (14400 * 1));

if ($check4leap == 1) {
$daysNyear="366";
}
else {
$daysNyear="365";
}

$doy = date(("z"), time() - (14400 * 1));
$calculate = $daysNyear - $doy;
$currentyear = date(("Y"), time() - (14400 * 1));
$output = $currentyear + 1;

if ($calculate == 0)
{
echo "Happy New Year";
}
else if ($calculate <= $daysNyear)
{
echo "There are $calculate days left until $output.";
}
?>

[dustinnoe]

Here is a more simplified solution

 

<?php
$doy = date(("z"), time() - (14400 * 1));
$currentYear = date(("Y"), time() - (14400 * 1));
$lastDay = strtotime("Dec 31, $currentYear");
$numDays  = date('z', $lastDay);
$daysTil = $numDays - $doy;

if ($doy == 0)
{
echo "Happy New Year!";
}
else
{
$newYear = $currentYear + 1;
echo "There are {$daysTil} days left until {$newYear}.";
}
?>