Re: Head Twister: Need Help Please

On Sat, 19 Apr 2003 20:23:31 -0400, you wrote: > I think I'm cross eyed after trying to write a script that would determine >the day of the year (date("z") and then do the math to get the number. If >you feel led to help out the fire department pleeeze do :-)... BTW, you >can see the old script below working at the top of www.svfd.org Here's a shot at a general solution, which should give you a hand. When you're trying to map a repeating seqeunce (your shift pattern) onto a longer seqeunce (the calender), mod (%) is your friend. My pattern counts days, but changing it to count weeks is pretty simple. There may be bugs (for a start, make sure you don't feed it calculate dates older than start dates) /* accepts a shift pattern as an array of text strings, and unix * timestamps denoting the date the shift pattern begins, and the * date which should be calculated for.returns the string that is * "active" on that day */ function calculateShift($shiftpattern, $starttime, $calctime) { /* calculate number of days between $starttime and $calctime */ $days = ((($calctime - $starttime) / 60) / 60) / 24; return $shiftpattern[(int)$days % sizeof($shiftpattern)]; } /* a shiftpattern array contains a series of text strings, * each denoting a shift. */ $joe = array("6-2", "6-2", "6-2", "off", "off", "2-10", "2-10", "2-10", "off", "off", "10-6", "10-6", "10-6", "off", "off"); $fred = array("2-10", "2-10", "2-10", "off", "off", "10-6", "10-6", "10-6", "off", "off", "6-2", "6-2", "6-2", "off", "off"); $r = calculateShift($joe, strtotime("01 April 2003"), strtotime("04 April 2003")); echo ("<p>On 4th April 2003, Joe will be on " . $r . " shift</p>"); $r = calculateShift($fred, strtotime("01 April 2003"), strtotime("08 April 2003")); echo ("<p>On 8th April 2003, Fred will be on " . $r . " shift</p>"); $r = calculateShift($fred, strtotime("01 April 2003"), strtotime("25 December 2003")); echo ("<p>On 25th December 2003, Fred will be on " . $r . " shift</p>");