Re: if(date("Y-m-d") >

Maybe this will help. Just keep adding to $var[] until you add all your weeks. Then execute it. This is written to run on the command line, but you can replace the \n with <br> to get your breaks in HTML. <?

       $var = Array();
       $var[] = "2005-01-02";
       $var[] = "2005-01-09";
       $var[] = "2005-01-16";

       $numweeks = count($var);
       echo "Number of weeks: " . $numweeks . "\n";

       $today = date("Y-m-d");
       for ($i=0;$i<$numweeks;$i++) {
               if ($today >= $var[$i]) {
                       $lastweek = $i + 1;
               }
       }
       echo "This is week $lastweek.\n";

?> HTH Trav www.pinkbunnysuit.com ----- Original Message ----- From: "John Taylor-Johnston" <taylorjo@collegesherbrooke.qc.ca> To: <php-general@lists.php.net> Sent: Monday, January 10, 2005 12:08 PM Subject: [PHP] if(date("Y-m-d") >

Hi, I would like some help to improve this script. I'm a teacher with a schedule of 17 weeks. Instead of using if(date("Y-m-d") >= $week3) I would like to do a "for i = 1 to 17" and if the current date date("Y-m-d") = week[i] I would like to echo "This is week $week[i]"; Can someone show me how please? <?php #old code: $week1 = "2005-01-17"; $week2 = "2005-01-24"; ... $week17 = "2005-05-09; if(date("Y-m-d") >= $week3) { echo "this is week 3"); } ?> Thanks, John -- John Taylor-Johnston ----------------------------------------------------------------------------- "If it's not Open Source, it's Murphy's Law."

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