RE: [PHP] Links with parameters in DB

[snip] It doesn´t work, I have other fields in the DB apart from the field 'Link', so if I use $link = eval($result); I get a parse error. Apart from that, I have to write the name of the field (link), if not the server won´t know the field I´m refering to. [/snip] Then did you eval that? I meant for you to use a proper rendering of the eval statement by itself, i.e. $sql = "SELECT * FROM table "; $result = mysql_query($sql, $connection); while($row = mysql_fetch_array($result)){ $link = eval($row['Link']); echo $link; }