Broomy

Ah, i'm good at this!  :-\ Thanks for that.

Thanks for the help, I appreciate it. That problem is now fixed but i'm not met with this error: Parse error: syntax error, unexpected '{' on the line that refers to this: if($row = db_fetch_result($result) { Any more help would be great, Thanks again.

Hi. I have this code that someone wrote for me, but it gives me an error if I try and use it as follows: Parse error: syntax error, unexpected T_VARIABLE The line with the error refers to $sql = " I'm useless at PHP but i'm hoping that someone might be able to point out what the problem might be. Thanks for any help. [code]<?php $currentNodeCreated = $node->created; $taxonomyTerm = 2 //this should not be hardcoded, it should be pulled from the node details for the current node taxonomy $sql = "   SELECT node.title, node.nid   FROM {node}, {term_node}   WHERE node.nid = term_node.nid   AND node.tid = $taxonomyTerm   AND node.created < $currentNodeCreated   AND node.status = 1   ORDER BY node.created DESC   LIMIT 1"; $result = db_query($sql); if($row = db_fetch_result($result) {   $output .= l($row->title, "node/".$row->nid); } ?>[/code]