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Posted by Joseph Melnick on 09/28/08 11:17
Hello Luigi,
What you could try: without using a database.
1. create a file with all of the picture names. named: pictures.txt
name1.jpg
name2.jpg
name3.jpg
2. use date() to give you the day of the year
$todayspicture = 'defaultpic.jpg';
$dayofyear = date("z"); // day of year 0-365
// today is day 149 so to start from zero subtract from 149 from $dayof year
$lines = file("pictures.txt");
if(count($lines)<$dayofyear) $todayspicture = $lines[$dayofyear];
Hope this helps
Joseph Melnick
JM Web Consultants
http://www.jphp.com
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