Posted by Tony on 10/10/49 11:17

"Joseph Melnick" <jmelnick@jphp.com> wrote in message
news:GYSdnSQYc4sl5QbfRVn-uA@rogers.com...
> 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

You could also read the filenames from the directory to populate the
filename array - although that would force everything into alphabetical
order..

>
> 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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