i tend to take the approach of
$next_wed = date("Y-m-d", strtotime("next wednesday"));
Bastien
From: Niel Archer <niel@xxxxxxxxxxxxx>
Reply-To: php-db@xxxxxxxxxxxxx
To: php-db@xxxxxxxxxxxxx
Subject: Re: SELECT date query
Date: Sat, 07 Oct 2006 05:49:36 +0100
Hi Ron
I've made the assumption that if today is Wednesday, you still want next
Wednesday.
Try this:
$offset = array(3,2,1,7,6,5,4);
$date = explode("-", date("Y-n-j"));
$ToDay = DayOfWeek($date[0], $date[1], $date[2]);
$NextWed = date("Y-n-j", time() + ($offset[$ToDay] * 24 * 60 * 60));
// Returns a digit in range 0-6. 0 = Sunday, 6 = Saturday
function DayOfWeek($Year, $Month, $Day)
{
$t = array(0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4);
$Year -= $Month < 3;
return ($Year + ($Year / 4) - ($Year / 100) + ($Year / 400) +
$t[$Month - 1] + $Day) % 7;
}
Niel
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