Man, you can't do echo Array!
you have to walk through an array!
dump the array and see its content via:
var_dump($retVal) for example,
refer yourself to the array functions in PHP's manual,
"Daniel alsén" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Hmm...
>
> it still doesn´t work.
>
> If i echo $retVal i just get the word 'Array' (the same number of times
that
> the number of files in the directory).
>
> If i echo $file i get the file listing...however, it doesn´t pass on the
> content to the rest of my script.
>
> Maybe someone could help me out with the whole script? I downloaded this
> from some script archive and it works if i use the list.photos datafile.
But
> i don´t want to change that file every time a new photo is uploaded.
> I am trying to read the directory and use the directory handle listing
> instead.
>
> If anyone wants to help me out making this databasedriven instead youre
very
> welcome :)
>
> <?PHP
>
> $retVal = array ();
> //Load Directory Into Array
> $handle=opendir('.');
> while (false !== ($file = readdir($handle)))
>
> if ($file != "." && $file != ".." && ereg(".jpg",$file)) {
>
> $retVal [count($retVal)] = $file;
>
> echo "$retVal <br>\n";
>
> }
>
> //initialize variables
> //$data_file = "list.photos";
> $thumbnail_dir = "thumbs/";
> $num_rows = 3;
> $photos_per_row = 3;
> $photos = file($retVal); //$retVal used to be $data_file
> $total_photos = sizeof($photos);
> $photos_per_page = $num_rows * $photos_per_row;
> //check to see if the start variable exists in the URL.
> //If not, then the user is on the first page - set start to 0
> if(!isSet($start)){
> $start = 0;
> }
> //init i to where it needs to start in the photos array
> $i = $start;
> $prev_start = $start - $photos_per_page;
> $next_start = $start + $photos_per_page;
>
>
> for ($row=0; $row < $num_rows; $row++){
> print("<tr>\n");
> for ($col=0; $col < $photos_per_row; $col++){
> if($i < $total_photos){
> $thumbnail = $thumbnail_dir.trim($photos[$i]);
> $thumb_image_size = getimagesize($thumbnail);
> $image_size = getimagesize(trim($photos[$i]));
> print("<td align=\"center\">
> <a
>
href=\"javascript:photo_open('photo_display.php?photo=".trim($photos[$i])."'
> ,'".$image_size[0]."','".$image_size[1]."');\"><img src=\"".$thumbnail."\"
> ".$thumb_image_size[3]." border=\"0\"></a></td>\n");
> } else {
> print("<td></td>\n");
> }
> $i++;
> }
> print("</tr>\n");
> }
>
>
> //Clean up directory array
> closedir($handle);
> return $retVal;
>
>
> //end table
> ?>
>
> > -----Original Message-----
> > From: _lallous [mailto:[EMAIL PROTECTED]]
> > Sent: den 18 september 2001 13:46
> > To: [EMAIL PROTECTED]
> > Subject: [PHP] Re: Replacing datafile with array
> >
> >
> > works like a charm....
> >
> > just initializet the $retVal function...
> >
> > $retVal = array();
> > rest of script here....
>
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