[ lordofblood @ 27.05.2010. 22:01 ] @
imam problem sa php i mysql upitom, dakle sve radi lepo sem komande koja treba da napravi pod-meni, proverao sam sql i tablea je dobra, vrednosti su dobre, sta vise mogu da ih ispisem posebno, ali nemogu da ih dodam kao pod-meni :( ako neko ima ideju ... bilo kakvu ... btw pocetnik sam tako da molim bez mnogo OMG :P Code: <?php $query = "SELECT * FROM subjects ORDER BY position ASC"; $subject_set = mysql_query($query, $connection); if(!$subject_set) { die("Database query faild!"); } while ($subject = mysql_fetch_array($subject_set)) { echo "<li>{$subject["menu_name"]}</li>"; } $query = "SELECT * FROM pages WHERE subject_id = {$subject["id"]} ORDER BY position ASC"; $page_set = mysql_query($query , $connection); if(!$page_set) { die("Database query faild!! ". mysql_error()); } echo "<ul id=\"pages\">"; while ($page = mysql_fetch_array($page_set)) { echo "<li>{$page["menu_name"]}</li>"; } echo "</ul>"; ?> .. izbacuje mi sledecu gresku Database query faild!! You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'ORDER BY position ASC' at line 4 |