Hi, A newbie question. I have more than one table to access from a database. When I use the code as below, it gives no response on the web page. What may I do to run more than one table from the same database into the script. BRgds, kNish . . . . <tr> <td height="33"> </td> <td width="14%" class="style3">Artist</td> <?php $options = mysql_query("SELECT artist_name FROM artist"); $options=mysql_fetch_array($options); echo '<td>'; enhanced_list_box(array( 'table' => 'artist', 'value_field' => 'artist_name')); function enhanced_list_box($options){ $sql = "select " . $options['value_field']; $sql .= " from " . $options['table']; $result = mysql_query($sql)or die("error in SQL"); echo '<select name="', $options['value_field'],'" size="1">'; while ($row = mysql_fetch_array($result, MYSQL_NUM)) echo '<option value="' . $row[0] . '">' . $row[0] . '</option>'; echo '</select>'; } echo '</td>'; ?> </tr> <tr> <td height="33"> </td> <td width="14%" class="style3">Project</td> <?php $options = mysql_query("SELECT project_name FROM project"); $options=mysql_fetch_array($options); echo '<td>'; enhanced_list_box(array( 'table' => 'project', 'value_field' => 'project_name')); function enhanced_list_box($options){ $sql = "select " . $options['value_field']; $sql .= " from " . $options['table']; $result = mysql_query($sql)or die("error in SQL"); echo '<select name="', $options['value_field'],'" size="1">'; while ($row = mysql_fetch_array($result, MYSQL_NUM)) echo '<option value="' . $row[0] . '">' . $row[0] . '</option>'; echo '</select>'; } echo '</td>'; ?> </tr> <http://www.php.net/a> . . . . .