Re: holding values in a select list on a form
| From: | Mark Bayfield | Date: | Mon, 23 Jul 2001 11:34:24 +0000 |
| Subject: | Re: holding values in a select list on a form | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-59179@lists.php.net to get a copy of this message | ||
I worked it out by doing something like this...
<?
if (!$WR_COUNTRY =="") { $topic = $WR_COUNTRY; };
echo "<select name=\"WR_COUNTRY\">";
$toplist = mysql_query("select * from wr_country where
wr_country_enabled = 1 ");
echo "<option value=\"\">Select</option>\n";
while(list($topicid, $topics) = mysql_fetch_row($toplist)) {
if ($topicid==$topic) {
$sel = "selected ";
}
echo "<option $sel value=\"$topicid\">$topics</option>\n";
$sel = "";
}
?>
"Mark Bayfield" <mbayfield@optushome.com.au> wrote in message
news:20010723093128.35422.qmail@pb1.pair.com...
> Some help please...
>
> I am creating a select list from a database, and I am trying to hold the
> value of what has been selected by a user, while I do some error checking.
> It is searching a mysql db to pull out the list. It will then need to pass
> values back into the database. The code I am using is this...
>
> <? echo "<select name=\"FIELDNAME\">";
> print "<option value=\"\">Select</option>";
> for ($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
>
>