Displaying images in a row problem.

[Sonic_Rage]

This is really driving me nutz now. I can't get the images to display 3 in a row, it keeps going down the list as 1 in a row. What am I doing incorrectly in my code? 😕

echo "<table width=100% bgcolor=#FFFFFF border=0 cellpadding=2 cellspacing=2>"; 

$query = "SELECT * FROM rpg_images WHERE gameid='12' ORDER BY gameid ASC limit $offset,$limit "; 
$result = mysql_query($query) or die("Could not execute query."); 
while ($row = mysql_fetch_array($result)) { 

// Make each row contain 3 or more images

$i=0;
  {
    { 
      if (!($i % 3))
        { 
           echo"<tr>";
        }
       $i++;
     }
     echo "<TD>";
     $url = "show_images.php?imageid=".($row["imageid"]);
     $title =  "<IMG SRC=\"images/".$row[img_thumb]."\" ALT=\"Click for larger image\" WIDTH=160 HEIGHT=120 BORDER=1>";
     do_html_url($url, $title);
     echo"</TD>";
  }
echo "</tr>";

// echo "<tr><td><IMG SRC=\"images/".$row[img_thumb]."\" BORDER=0></td></tr>"; 

} 
mysql_free_result($result); 


echo "</table>"; 

[dc2_jspec]

looks to me like you are re-setting $i=0 everytime you go through the while loop, maybe the below code will help a little.

echo "<table width=100% bgcolor=#FFFFFF border=0 cellpadding=2 cellspacing=2>";

$query = "SELECT * FROM rpg_images WHERE gameid='12' ORDER BY gameid ASC limit $offset,$limit ";
$result = mysql_query($query) or die("Could not execute query.");

$i=1;
echo "<tr>";
while ($row = mysql_fetch_array($result))
{
if ($i%4==0)
{
echo "</tr>
echo"<tr>";
}
echo "<TD>";
$url = "show_images.php?imageid=".($row["imageid"]);
$title = "<IMG SRC=\"images/".$row[img_thumb]."\" ALT=\"Click for larger image\" WIDTH=160 HEIGHT=120 BORDER=1>";
do_html_url($url, $title);
echo"</TD>";

$i++;
// echo "<tr><td><IMG SRC=\"images/".$row[img_thumb]."\" BORDER=0></td></tr>";

}
echo "</tr>";
mysql_free_result($result);

echo "</table>";