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@NogDogDec 11.2008 — #I think you may be neglecting the fact that this is a leap year, and as such there are 366 days this year, so 345 is correct for Dec. 11, 2008. @NogDogDec 12.2008 — #OK, so if we start at 0, that means this year (a leap year) Dec. 31 will be day number 365 (The 366th day of the year minus 1). If we subtract 20 from 365 we get 345, and if we subtract 20 from Dec. 31 we get Dec. 11. Therefore, the "z" day for Dec. 11, 2008 is 345, which is what I get with [b]date('z');[/b].
Output:
[RESOLVED] PHP date function ‘z’
Hello using the PHP date function “z” says it goes from 0-365 found here [url]http://www.php.net/manual/en/function.date.php
But 0 shows is a non working field why do they list it that way? I can only get something to actually show
for today dec 11 2008 [url]http://www.vpcalendar.net/Julian_Date.html
When according to the php manual should only show at 344 instead since first one is supposed to start at 0 not 1.
Can someone please explain this to me? Am trying to explain it to someone else on how php manual yearly date z works and can’t, since it isn’t making any sense.
Please let me know why the php manual says it should start at 0 when items are only showing starting at 1 instead. Thank you very much for your help.
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@NogDogDec 11.2008 — #0
@NogDogDec 12.2008 — #OK, so if we start at 0, that means this year (a leap year) Dec. 31 will be day number 365 (The 366th day of the year minus 1). If we subtract 20 from 365 we get 345, and if we subtract 20 from Dec. 31 we get Dec. 11. Therefore, the "z" day for Dec. 11, 2008 is 345, which is what I get with [b]date('z');[/b].[code=php]
<?php
$dates = array('2008-01-01', '2008-12-11', '2008-12-31');
foreach ($dates as $date)
{
echo "<p>$date: ".date('z', strtotime($date))."</p>n";
}[/code]Output:
<i>
</i>2008-01-01: 0
2008-12-11: 345
2008-12-31: 365