Hi.
I am building a very small and simple photo gallery, that will fetch an images url and a comment assigned to this image, from a MySQL database.
Here is a code snippet that causes some errors:
(yes, i know the coding might not be the best, and i appreciate any suggestions on how to make it better..)
if($_GET['sesjon'] == "")
{
$q = mysql_query("select * from directories");
$n = mysql_num_rows($q);
for($i=0; $i < $n; $i++)
{
$r = mysql_fetch_array($q);
echo "<a href = \"?side=bilder&sesjon=".$r["path"]."\">".stripslashes($r["title"])."</a><br>";
}
}
else{
$per_row = 2; // Number of images per row
$x = 1; // Counter variable
echo '<table width="534" border="0" cellspacing="1" cellpadding="2"><tr>';
$q = mysql_query("select * from photos where category = '$sesjon'");
$n = mysql_num_rows($q);
$r = mysql_fetch_array($q);
while ($r = mysql_fetch_array($q))
{
$filecomment = stripslashes($r["comments"]);
$bildeuri = $r['category'].$r['photo'];
$source = @imagecreatefromjpeg($bildeuri);
$y = (imageSY($source))/2;
$x = (imageSX($source))/2;
@imagedestroy($source);
echo "<td class=\"bodytext\"><img src = ".$bildeuri." height = ".$y." width =".$x."><br>".$filecomment."</td>";
if ( $x == (int)$per_row )
{
echo "</tr>";
$x = 1;
}
else
{
$x++;
}
}
echo '</table>'; } ?>
And these are the errors that i get:
Warning: imagesy(): supplied argument is not a valid Image resource in xxxxxxxx/bilder.php on line 45
Warning: imagesx(): supplied argument is not a valid Image resource in xxxxxxxx/bilder.php on line 46
AND also:
All the images are printed one one row.. It should break on every second photo..