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

> On Mon, 10 Jan 2005 13:08:28 -0500, John Taylor-Johnston > <taylorjo@collegesherbrooke.qc.ca> wrote: >> 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? You could also consider using http://php.net/date with the 'W' (ISO-8601 week number of year, weeks starting on Monday (added in PHP 4.1.0)) argument or 'z' (The day of the year (starting from 0)) and divide by 7. You'd need to compute your offset to your 'first' week that you care about, but... <?php //PHP 4.1.0 and up: $start_week = date('W', mktime(0, 0, 0, 9, 4, 2005)); echo "This is week: ", date('W', mktime(0, 0, 0, 10, 1, 2005)) - $start_week, "<BR>\n"; ?> Depending on what you are doing and how you are already storing dates, this could be cleaner. Watch out, though, as you'll have issues on New Year's Eve if that's in the middle of a 17-week period, as I suspect it is... You could also just use: <?php $start_week = mktime(0, 0, 0, 8, 2, 2005); function seconds_to_week($seconds){ return $seconds / (60*60*24*7); } echo "This is week: ", seconds_to_week(mktime(0, 0, 0, 9, 17, 2005)), "<BR>\n"; ?> -- Like Music? http://l-i-e.com/artists.htm