Making menu with submenu

[Bravat]

I am trying to make menu with submenu. I have two tables:

meni(id_meni, naslov, pozicija, lang_id) and

subjects (subjecte_id, naslov, text, meni_id, lang_id, pozicija).

Into subjects(meni_id) is written meni(id_meni) value in order to determinate what menu items are having summenu options. How can i do this? I tried following code, but it only display items from meni table:

<?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>"; 
			 }
			?>

[yonny]

Hi, I have done these menus before and I think you need to start with a <ul> before any while loop. While inside the first loop for the fist menu print <li> link </li> then if the second sql query has a result the second <ul> that starts the submenu needs to print before the second while loop. The second while loop needs to print only <li> link </li>. once that loop is done close the second </ul> ending the submenu.

After you closed the first loop then close </li> and </ul>.

 

 

[attachment deleted by admin]

[yonny]

this is the code you posted on another post

<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>

so to make it work with what you have

 

// make your sql_query
// count the affected rows
// if not 0 start the menu by printing <ul>

// start your first loop
print "<li><a href=\"$some_link\">$some_column</a>"; \\ do not close <li> yet because you dont know if it has a submenu

// do the second sql query
// count the affected rows
// if 0 rows then 
print "</li>"; \\ to close that one and move on with the next

// else  (not 0 rows) then
print "<ul class=\"subnav\">";

// now run your second while loop
print"<li><a href=\"$sub_page\">$submenu_item</a></li>";
                           //etc  <li><a href="#">Small Business</a></li>
                            <li><a href="#">HTML Templates</a></li>
                            <li><a href="#">Joomla Templates</a></li> //
endwhile;
                print"</ul>";
endif;
          print"</li>";

//and repeat

 

if for some reason you need to link to anchors in the same page you need to set the mainmenu link to "#" as opposed to "website_file.php#"

and the submenues to "#anchor" instead of "website_file.php#anchor" or they will not work. i used PHP_SELF  to check if I was inside the page or not. so if the page I was in was about.php and i had anchors in it i would

print"<li><a href=\"#123\">$submenu_item</a></li>"; if it was not about.php i would

print"<li><a href=\"about.php#123\">$submenu_item</a></li>";

[Bravat]

I think that I am doing something wrong here, but I cannot see error. I managed to create submenu for specific menu, but it only shows one item in subemnu. Here is the code:

<?php
                 $query_meni =mysql_query("SELECT * FROM meni  WHERE lang_id = '$jezik'");
			if(mysql_affected_rows() >0 ){
			 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>";
				 $query_subject =mysql_query("SELECT * FROM subjects WHERE meni_id = '$row_meni[id_meni]' AND lang_id = '$jezik' ORDER BY pozicija ASC");	
						if(mysql_affected_rows() == 0){
							echo "<li>";}
							else {
							while($row_subjects = mysql_fetch_array($query_subject)){
							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>"; 
			 }
			?>

[yonny]

try this. i have not tested it but...

<?php
     $query_meni = "SELECT * FROM meni  WHERE lang_id = '$jezik'";
     $result = mysql_query($query_meni);
     $count = mysq_affected_rows();
	if($count >0 ):
	print"<ul class=\"main_menu\">"; /// open main menu
		while($row_meni = mysql_fetch_array($query_meni)):
		$link = $row_meni['id_meni'];
		$show = $row_meni['naslov'];
		print"<li><a href=\"index.php?page=$link\">$show</a>";

$query_subject = "SELECT * FROM subjects WHERE meni_id = '$row_meni[id_meni]' AND lang_id = '$jezik' ORDER BY pozicija ASC";	
        $result1 = mysql_query($query_subject);
$count1 = mysql_affected_rows();

		if($count1 == 0):
		print "</li>"; /// close list
			else:
			print"<ul class=\"subnav\">"; /// start sublist before 2nd while loop
					while($row_subjects = mysql_fetch_array($query_subject)):
					$link1 = $row_subjects['subject_id'];
					$show1 = $row_subjects['naslov'];
					print"<li> <a href=\"index.php?subject=$link1\">$show1</a></li>";
					endwhile;
					print"</ul>"; /// end sublist after while loop
			endif;
			print"</li>"; /// close list at endif
		endwhile;				
		print"</li>"; ///close parent
	endif;
	print"</ul>";  /// close main
?>

[Bravat]

It is working perfectly  . Thank you a lot for helping me  .

[yonny]

You are welcome. I'm glad it worked out.