Randy + Chris:
Thx for the quick replies!
Chris:
$rows contain ip addresses (postgresql inet) type (example:
192.168.0.1)
$options contain ip addresses (postgresql inet0 type (example:
127.0.0.1)
(keep in mind) they are both arrays. $rows contact all ips, and $options contact the ip addresses that were selected
Randy:
I am not sure what you mean. The variables look right. I am getting no errors! Just the logic is messed up.
HTH
On 12/3/06, Randy Moller <zoomerz@comcast.net> wrote: Maybe it's just a typo, but your pg_query is returning into the var
$result_ip, while your fetch is referencing var $result (without _ip).
Maybe that's just a typo for your question, but if not, that is at least one
error.
Also, it would be helpful to know what error you're getting....
Randy Moller
-----Original Message-----
From: pgsql-php-owner@postgresql.org [mailto:pgsql-php-owner@postgresql.org]
On Behalf Of Chris
Sent: Sunday, December 03, 2006 4:08 PM
To: Mag Gam
Cc: pgsql-php@postgresql.org
Subject: Re: [PHP] HTML FORMS selected question
Mag Gam wrote:
> I am having trouble getting values populated in my form... Here is what
> I got so far:
>
> //The values that need to be selected
> $result_ip=pg_query($dbconn,$test_query);
>
> <SELECT NAME="ip[]" SIZE=10 MULTIPLE>
> <?php
> //This will show ALL values.
> while($rows=pg_fetch_array($result)){
> foreach ($options as $index => $option) {
> if ($rows[1]==$option)
> {
What does $rows contain? What does $options contain?
You could also simplify this a little bit. I'd at least remove the
foreach $options loop and use "in_array" - see php.net/in_array
while ($rows = pg_fetch_array($result)) {
$selected = '';
if (in_array($rows[1], $options)) {
$selected = ' SELECTED';
}
sprintf('<option value="%s"%s>%s</option>', $rows[1], $selected,
$rows[1]);
}
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