如何使用php创建下拉菜单？

Question

This title may confuse someone, but the my goal is to make a dropdown menu with echoing it within the PHP function. To understand my problem a little bit...

My mainpages, the names are displayed at the top menu:

TABLE INDEX_PAGES

id,    name,        href
----------------------------
1     login     ?page=login    
2     admin     ?page=admin  
3     projects  ?page=projects   
4     home      ?page=home  

The table to know what mainpages can be displayed in the top menu

TABLE TOP_MENU

id,    index_page
------------------
1     1        
2     2       
3     3  
4     4        

Im using function to echo them into my topmenu.

Structure:

<header>
<h1 class="header">Godhaze</h1>
<nav id="nav-vasak">
<ul>
<?php echo get_topnav(); ?>
</ul>
</nav>
</header>

Function:

function get_topnav(){
global $db;
$sql="SELECT i.name,i.href from index_pages i join top_menu t on t.index_page=i.id ";
foreach($db->query($sql) as $row) {
echo  "<li><a href='".get_domain().$row['href']."'>".$row['name']."</a>";
//dropdownmenu
echo "</li>";
}
}

Now I have topmenu names and its all working fine, but I want to display my second pages in dropmenu view.

Like:

<header>
    <h1 class="header">Godhaze</h1>
    <nav id="nav-vasak">
    <ul>
    <li><a>Login</a> //mainpage
    <li><a>Action_log</a></li> //secondpage
   </li>
    </ul>
    </nav>
    </header>

TABLE SECOND_PAGES

id,    name,     index_page_id,                href
------------------------------------------------------------------
1     error_log       2           ?page=admin&secondpage=error_log    
2     user_log        2           ?page=admin&secondpage=user_log
3     action_log      1          ?page=login&secondpage=action_log
4     test_log        3           ?page=projects&secondpage=test_log 

The table to know what secondpages can be displayed in the dropdown top menu, index_page is the foreign key to index_pages and second_page is foreign key to second_pages

TABLE TOP_MENU_PAGES

id,    index_page, second_page
-------------------------------
1          1           3   
2          2           1

So now to my question is.. how can I make this function so I can display the second pages at dropdown menu?

Answer 1

Try this modified function

function get_second_pages(){
 global $db;
 $sql="SELECT i.name,i.href,s.name as name2,s.href as href2,
       from index_pages i 
       join top_menu t on t.index_page=i.id 
       join SECOND_PAGES s on t.second_page=s.id ";
 foreach($db->query($sql) as $row) {
  echo  "<li><a href='".get_domain().$row['href']."'>".$row['name']."</a>";
  echo  "<li><a href='".get_domain().$row['href2']."'>".$row['name2']."</a>";
  echo "</li>";
 }
}

I thing it will help you.