Bravat Posted March 25, 2011 Share Posted March 25, 2011 I want to create menu with submenu. Here is HTML code: <!--<li><a href="#">Home</a></li> <li><a href="#">About Us</a></li> <li><a href="#">Web Design</a> <ul class="subnav"> <li><a href="#">Corporate</a></li> <li><a href="#">Small Business</a></li> <li><a href="#">HTML Templates</a></li> <li><a href="#">Joomla Templates</a></li> </ul> </li> <li><a href="#">Web Development</a> <ul class="subnav"> <li><a href="#">eCommerce</a></li> <li><a href="#">CMS</a></li> <li><a href="#">Custom Features</a></li> <li><a href="#">Web Maintenance</a></li> </ul> </li> <li><a href="#">Internet Marketing</a> <ul class="subnav"> <li><a href="#">SEO</a></li> <li><a href="#">Pay Per Click</a></li> <li><a href="#">Social Media Marketing</a></li> <li><a href="#">Link Building</a></li> <li><a href="#">Content Marketing</a></li> <li><a href="#">Reputation Management</a></li> </ul> </li> <li><a href="#">Sound</a></li> <li><a href="#">3D</a></li> <li><a href="#">Portfolio</a></li> <li><a href="#">Quote</a></li> <li><a href="#">Contact</a></li>--> and this is php code which is trying to imitate HTML: <?php $query_meni =mysql_query("SELECT * FROM meni WHERE lang_id = '$jezik'"); $query_subject =mysql_query("SELECT * FROM subjects WHERE lang_id = '$jezik'"); while($row_meni = mysql_fetch_array($query_meni)){ echo "<li>"; echo "<a href=\"index.php?page="; echo $row_meni['id_meni']; echo "\">"; echo $row_meni['naslov']; echo "</a>"; while($row_subjects = mysql_fetch_array($query_subject)){ if($row_meni['id_meni'] == $row_subjects['meni_id']){ echo "<ul class=\"subnav\">"; echo "<li>"; echo "<a href=\"index.php?subject="; echo $row_subjects['subject_id']; echo "\">"; echo $row_subjects['naslov']; echo "</li>"; echo "</a>"; echo "</ul>"; } } echo "</li>"; } ?> Any idea why this is not working? Quote Link to comment Share on other sites More sharing options...
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