I am really not sure what kind of problem this is but my guss is that it has to do something with php here is the code:
echo "<tr>
<td colspan=\"3\" align=\"center\">";
for ($I=0;$I<$nrow;$I++)
{
$row1 = mysql_fetch_array($result);
extract($row1);
echo "<a href=\"./result?result_catalog=$TBL\"><img border=\"0\" src=\"http://www.eezs.com/images/$Picture\" width=\"111\" height=\"90\" /></a>";
if ($I == 4)
{
echo "</tr>
<tr>
<td colspan=\"3\" align=\"center\">";
}
elseif ($I == 8)
{
echo "</tr>
<tr>
<td colspan=\"3\" align=\"center\">";
}
elseif ($I == 12)
{
echo "</tr>
<tr>
<td colspan=\"3\" align=\"center\">";
}
elseif ($I == 16)
{
echo "</tr>
<tr>
<td colspan=\"3\" align=\"center\">";
}
else
{
echo " ";
}
}
echo "</tr>";
The problem is that when ever it displays the picture with link the picture is one for all the diffrent pictures that were suppose to display and same goes with the link so is it something wrong with the php or what also there is something above this code too and it is :
$query = "SELECT * FROM `Catalog` WHERE `ED` = 'E'";
$result = mysql_query($query)
or die("DB Error Please try again later");
$nrow = mysql_num_rows($result);
//to echo out links
echo "<tr align=\"center\" valign=\"top\">
<td colspan=\"3\"><font color=\"#008000\" face=\"Arial\">Categories</font>";
for ($i=0;$i<$nrow;$i++)
{
$row = mysql_fetch_array($result);
extract($row);
echo "<br />
<a href='./result?result_catalog=$TBL'><font face=\"Arial\" size=\"2\" color=\"#969493\"><u>$Name</u></a>";
}
echo "</font><br /></td></tr>";
I am gussing there could be something wrong here too.w