https://interactivetools.com/forum/forum-posts.php?Problem-with-Double-WHERE-statement-71708 Mon, 15 Jun 2026 20:10:40 -0700 en-us https://interactivetools.com/forum/forum-posts.php?postNum=2206644#post2206644
The where statement has to be a string. Try this:

<?php
$where="1";

if(!$CURRENT_USER['staff']){
$where = "conflicts_of_interest NOT LIKE '%\t".intval($CURRENT_USER['num'])."\t%'";
}

list($applicationsRecords, $applicationsMetaData) = getRecords(array(
'tableName' => 'applications2010',
'where' => "$where AND full=1",
));

?>

You'll want to use $where="1" because if the current user is not staff, you'll have an "AND" in your where clause with no initial where statement. This would result in an error.

Hope this helps
]]> Wed, 08 Sep 2010 17:14:21 -0700 forum-posts.php?postNum=2206644#post2206644 https://interactivetools.com/forum/forum-posts.php?postNum=2206636#post2206636
<?php
$where="";

if(!$CURRENT_USER['staff']){
$where = "conflicts_of_interest NOT LIKE '%\t".intval($CURRENT_USER['num'])."\t%'";
}

list($applicationsRecords, $applicationsMetaData) = getRecords(array(
'tableName' => 'applications2010',
'where' => $where AND 'full=1',
));

?>
]]> Wed, 08 Sep 2010 16:20:50 -0700 forum-posts.php?postNum=2206636#post2206636