hellonoko Posted December 21, 2008 Share Posted December 21, 2008 I am new to XML and trying to do a RSS tutorial. I am receiving the following error, </channel> is almost at the bottom: XML Parsing Error: not well-formed Location: http://www.sharingizcaring.com/rss/ Line Number 1, Column 3: </channel> ---------^ Code: <? class RSS { public function RSS() { require_once ('mysql_connect.php'); } public function GetFeed() { return $this->getDetails() . $this->getItems(); } private function dbConnect() { DEFINE ('LINK', mysql_connect (DB_HOST, DB_USER, DB_PASSWORD)); } private function getDetails() { $detailsTable = "webref_rss_details"; $this->dbConnect($detailsTable); $query = "SELECT * FROM ". $detailsTable; $result = mysql_db_query (DB_NAME, $query, LINK); while($row = mysql_fetch_array($result)) { $details = '<?xml version="1.0" encoding="ISO-8859-1" ?> <rss version="2.0"> <channel> <title>'. $row['title'] .'</title> <link>'. $row['link'] .'</link> <description>'. $row['description'] .'</description> <language>'. $row['language'] .'</language> <image> <title>'. $row['image_title'] .'</title> <url>'. $row['image_url'] .'</url> <link>'. $row['image_link'] .'</link> <width>'. $row['image_width'] .'</width> <height>'. $row['image_height'] .'</height> </image>'; } return $details; } private function getItems() { $itemsTable = "webref_rss_items"; $this->dbConnect($itemsTable); $query = "SELECT * FROM ". $itemsTable; $result = mysql_db_query (DB_NAME, $query, LINK); $items = ''; while($row = mysql_fetch_array($result)) { $items .= '<item> <title>'. $row["title"] .'</title> <link>'. $row["link"] .'</link> <description><![CDATA['. $row["description"] .']]></description> </item>'; } $items .= '</channel> </rss>'; return $items; } } ?> Quote Link to comment Share on other sites More sharing options...
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