Code-Nummer mit mysql Multidimensional Dynamic Menu Creation?

Question

Hallo Ich entwickle eine Website mit Codeigniter mit mehrdimensionalen dynamischen Menüs. Ich benutze Codeingiter und mysql Ich habe create Menü, submenu1, submenu2. Jetzt wird das Menu und Submenu1 in die Webseite geladen, aber ich versuche submenu2 zu laden. Ich bekomme das Problem, dass ich die array_push Methode in PHP verwende. Hier ist mein Code in diesem Code i bereits Lademenü und submenu1 kann jede Stelle mir helfen, wie auch laden submenu2 ...

Dies ist mein Code anzeigen ..

<div class="collapse navbar-collapse"> 
        <ul class="nav navbar-nav"> 
         <li class="active"><a href="<?php echo base_url()?>Pixel/index" >Home</a></li> 
         <?php foreach($menus as $menu) { ?> 
          <li id="mainmenu"> 
           <a href="#" class="dropdown-toggle" id="<?php echo $menu->m_id; ?>" data-toggle="dropdown"> <?php echo $menu->menuname; ?> <b class="caret"></b></a> 
           <ul class="dropdown-menu"> 
            <?php 
            if (isset($menu->children)) { 
            foreach ($menu->children as $child) { 
             ?> 
            <li> 
             <a href="<?php echo base_url()?>Pixel/corporate" id="submenuone" class="dropdown-toggle"><?php echo $child->submenu_name; ?> 
             </a> 
            </li> 
            <?php 
             } 

            } ?> 
           </ul> 
          </li> 
         <?php } ?> 

Dies ist mein Controller-Code

public function index() 
    { 
       $menus = $this->Px_Model->populateMenus(); 
       $data = array('menus' => $menus); 
       $this->load->view('pxHead'); 
       $this->load->view('pxHeader'); 
       $this->load->view('pxNav',$data); 
       $this->load->view('pxSlider'); 
       $this->load->view('pxHome'); 
       $this->load->view('pxFooter'); 
    } 

Das ist mein Modellcode

public function populateMenus(){ 
    $this->db->select("*"); 
    $this->db->from("menu"); 
    $q = $this->db->get(); 
    $final = array(); 
    if ($q->num_rows() > 0) { 
    foreach ($q->result() as $row){ 
      $this->db->select("*"); 
      $this->db->from("submenu1"); 
      $this->db->where("menu_id", $row->m_id); 
      $q = $this->db->get(); 
      if ($q->num_rows() > 0) { 
       $row->children = $q->result(); 
      } 
      array_push($final, $row); 
     } 
    } 
    return $final; 
    } 

Answer 1

wie this.Need Versuchen mehrdimensionales Array zu machen.

MODEL

// $final = array(); 
if ($q->num_rows() > 0) { 
foreach ($q->result() as $row){ 
     $final[]['menuname'] = $row->menuname; 
     $final[]['m_id'] = $row->m_id; 
     $this->db->select("*"); 
     $this->db->from("submenu1"); 
     $this->db->where("menu_id", $row->m_id); 
     $q = $this->db->get(); 
     if ($q->num_rows() > 0) { 
      $final[]['children'] = $q->result(); 
     } 

    } 
} 
return $final; 
} 

VIEW

<div class="collapse navbar-collapse"> 
       <ul class="nav navbar-nav"> 
        <li class="active"><a href="<?php echo base_url()?>Pixel/index" >Home</a></li> 
        <?php foreach($menus as $key=>$menu) { ?> 
         <li id="mainmenu"> 
          <a href="#" class="dropdown-toggle" id="<?php echo $menu['m_id']; ?>" data-toggle="dropdown"> <?php echo $menu['menuname']; ?> <b class="caret"></b></a> 
          <ul class="dropdown-menu"> 
           <?php 
           if (isset($menu['children'])) { 
           foreach ($menu['children'] as $child) { 
            ?> 
           <li> 
            <a href="<?php echo base_url()?>Pixel/corporate" id="submenuone" class="dropdown-toggle"><?php echo $child->submenu_name; ?> 
            </a> 
           </li> 
           <?php 
            } 

           } ?> 
          </ul> 
         </li> 
        <?php } ?>