Image display mySQL

Anon

Thanks for any help in advance....I have a script that uploads image files into a directory, no problem there. Now that they are in that directory how to i pull them out using a loop to show everything the table, such as this one ...

for ($i = 0; $i < $rows; $i++) {
$data = mysql_fetch_object($result);
echo " <td>$data->image_name</td>\n";

echo " <td>$data->name</td>\n";
echo " <td>$data->description</td>\n";
echo " <td>$data->price</td>\n";
echo " </tr>\n";

Where image_name obvious refers to the name of the image stored in the database. Thanks Dave

tyrnt

If i understand you correctly you are doing something similar to what I am doing and you can use something similar to this

// Your Query
$resultID = mysql_query("SELECT * FROM items WHERE name LIKE '%$nameofitem%'", $link);

//tells how many rows and puts in $total_items
$total_items = mysql_num_rows($resultID);

// For loop parsing the table
for($x = 1; $x <= mysql_num_rows($resultID); $x++)
{
print ( "");

// putting in array
$row = mysql_fetch_assoc($resultID);
print "<tr>\n";
print "<td>" . $row[item_id] . "</td>\n";
print "<td><a href=\"viewitem.php3?item_id=" . $row[item_id] . "\"><b>" . $row[name] . "</b>\n";
print "<td>" . $row[type] . "</td>\n";
print "<td>" . $row[classification] . "</td>\n";
print "</tr>\n";
}
print "</table>\n";

$row[<field name here>] will get the value of that field name in order of your key id i believe. and you can use the variable within the html as you see above.

Hope this helps,

Tyrnt