<?      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:

> -----Original Message-----
> From: Mark Bayfield [mailto:[EMAIL PROTECTED]]
> Sent: 23 July 2001 10:34
> 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

PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]

Reply via email to