RE: holding values in a select list on a form

<? echo "<select name='FIELDNAME'>"; print "<option value=''>Select</option>"; if($result = mysql_db_query($db, $query)) while($row = mysql_fetch_row($result)) print "<option value={$row[1]}>{$row[1]}</option>"; echo "</select>"; ?> although personally I use mysql_fetch_array() so I can use the field name to get the appropriate field rather than getting the second field returned from the query (as you are doing). Tim Ward Senior Systems Engineer Please refer to the following disclaimer in respect of this message: http://www.stivesdirect.com/e-mail-disclaimer.html > -----Original Message----- > From: Mark Bayfield [mailto:mbayfield@optushome.com.au] > Sent: 23 July 2001 10:34 > To: php-general@lists.php.net > Subject: holding values in a select list on a form > > > Some help please... > > I am creating a select list from a database, and I am trying > to hold the > value of what has been selected by a user, while I do some > error checking. > It is searching a mysql db to pull out the list. It will then > need to pass > values back into the database. The code I am using is this... > > <? echo "<select name=\"FIELDNAME\">"; > print "<option value=\"\">Select</option>"; > for ($index = 0; $index < mysql_num_rows($query); $index++) { > $row = mysql_fetch_row ($query) or die (mysql_error()); > print "<option value=$row[1]>$row[1]</option>"; > } > echo "</select>"; > ?> > > If there is an easier way, let me know.... > > Mark > > >