Posted by Jeff Gardner on 09/15/06 06:32

Greetings:

I am attempting to get conditional output based on POSTed form data.
If the posted value is either the key or value of an array, $x=key and
$q=foo. elseif it is neither key nor value(I tried || too) , $q=bar.
Even though $x is properly assigned the new value, $q always ends up
being bar. Abstractions aside, here's the code: (note ## comments
added to this post)

<snip>
$search = $_POST[search];
$sort = $_POST[sortby];
$states = array(
"AL" => "Alabama",
"AK" => "Alaska",
"AZ" => "Arizona",
...
);
##echo $search here outputs the POSTed input
foreach ($states as $key => $value)

if ($search == $value || $search == $key) {
$search = $key;
##echo $search here outputs the new value, when applicable
$q = "
SELECT
organization.*, contacts.*
FROM organization
LEFT JOIN contacts
ON organization.org_id = contacts.org_id
OR organization.org_id = 0
WHERE organization.state
= '$search'
OR contacts.state
= '$search'
ORDER BY '$sort' ASC";

} elseif ($search != $value && $search != $key) { ##I tried || here too.
##echo $search here outputs the new value too, when applicable
$q = "
SELECT
CONCAT(contacts.firstname,' ',contacts.lastname) AS fullname,
organization.*, contacts.*
FROM organization
LEFT JOIN contacts
ON organization.org_id = contacts.org_id
OR organization.org_id = 0
WHERE organization.state
LIKE '%$search%'
OR organization.orgname
LIKE '%$search%'
OR contacts.state
LIKE '%$search%'
OR contacts.lastname
LIKE '%$search%'
OR contacts.firstname
LIKE '%$search%'
ORDER BY '$sort' ASC";
}

$result = mysql_query($q)
or die('Query Failed: ' . mysql_error());
</snip>

Where is this evaluation breaking down and why? Thanks in advance.
--

Regards,

Jeff Gardner
___________________________

"Contrary to popular belief, Unix is user friendly. It just happens
to be very selective about who its friends are." --Kyle Hearn

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