You'd probably be able to get the first Friday in a year with smth like:
$Friday = strtotime("First Friday in 2003"); And if you go for the "count 7
days up" method .. Pitfall could be forgetting to check if you're still in
the year you started in.

Hope I added something valueable ;)

-----Original Message-----
From: Martin Towell [mailto:[EMAIL PROTECTED] 
Sent: Friday, October 03, 2003 3:18 AM
To: 'J Morton'; Shew; [EMAIL PROTECTED]
Subject: RE: [PHP] date brain teaser

How about something similar, but look for the first Friday and then add
seven days (or 604800 seconds) to get the date of the next Friday?

-----Original Message-----
From: J Morton [mailto:[EMAIL PROTECTED]
Sent: Friday, 3 October 2003 11:20 AM
To: Shew; [EMAIL PROTECTED]
Subject: Re: [PHP] date brain teaser


Sure --

<?
$this_date = "1/1/2003";

for($i=1; $i<=365; $i++) {
    if (date("l", strtotime($this_date) == "Friday")) {
        echo $this_date . "<br>";
    }

    $this_date = date("n/j/Y",mktime(0,0,0,date("m",
strtotime($this_date)),date("d", strtotime($this_date))+1,date("Y",
strtotime($this_date))));
}
?>


Shew wrote:

> Hi,
>
> I'm trying to find a formula for displaying the date for every Friday of a
> given year, i.e. 2003
>
> Jan 3, 2003
> Jan 10, 2003
> Jan 17, 2003
> etc.
>
> Any ideas?
>
> Thanks
>
> --
> PHP General Mailing List (http://www.php.net/)
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