Question

我一直在尝试使用Codeigniter找到在数据库中创建动态导航菜单存储的解决方案。但到目前为止我无法解决我的问题，我真的不明白如何创建此功能我尝试创建下面的代码，但我只能显示没有子菜单的菜单。在视图中，我不仅可以选择菜单，还可以选择子菜单。我真的不明白如何为创建子菜单创建条件因为我必须将父与菜单 ID 进行比较，如果我发现父等于菜单 ID ，我会显示下拉当前该菜单但这对我来说是失败的

以下是我的观点

<div class="navbar-collapse collapse">
    <ul class="nav navbar-nav">
       <li><a>some menu retrieve from DB </a></li>
           <li class="dropdown megamenu-fullwidth"> 
              <a data-toggle="dropdown" class="dropdown-toggle">title</a>
              <ul class="dropdown-menu"> 
                <li class="megamenu-content">
                 <ul class="col-lg-2 col-sm-2 col-md-2">
                   <li class="no-border"><p><strong>Head</strong></p></li>
                   <li><a href="#">Descript<br></a></li>
                 </ul> 
               </li> 
            </ul>
        </li>
       <?PHP endif; ?>
      <?PHP endforeach; ?>
 </ul>    
</div>

这是控制器

$this->data['menus'] =  $this->nav_m->menus();

以下是模型

 public function menus() {
        $this->db->select("*");
        $this->db->from("nav"); 
        $this->q = $this->db->get();
        if ($this->q->num_rows() > 0) {
           return $this->q->result_array();
        }
    }

这是我在DB

中使用Var_dump（）的导航

array (size=7)
  0 => 
    array (size=9)
      'id' => string '1' (length=1)
      'uid' => string '0' (length=1)
      'text' => string 'title1' (length=6)
      'link' => string 'menus1' (length=6)
      'show_condition' => string '1' (length=1)
      'parent' => string '0' (length=1)
      'class' => string '' (length=0)
      'data' => string '' (length=0)
      'des' => string '' (length=0)
  1 => 
    array (size=9)
      'id' => string '2' (length=1)
      'uid' => string '0' (length=1)
      'text' => string 'title2' (length=6)
      'link' => string 'menus2' (length=6)
      'show_condition' => string '1' (length=1)
      'parent' => string '2' (length=1)
      'class' => string '1' (length=1)
      'data' => string '' (length=0)
      'des' => string '' (length=0)
  2 => 
    array (size=9)
      'id' => string '3' (length=1)
      'uid' => string '0' (length=1)
      'text' => string 'title3' (length=6)
      'link' => string 'menu3' (length=5)
      'show_condition' => string '1' (length=1)
      'parent' => string '2' (length=1)
      'class' => string '' (length=0)
      'data' => string '' (length=0)
      'des' => string '' (length=0)
  3 => 
    array (size=9)
      'id' => string '4' (length=1)
      'uid' => string '0' (length=1)
      'text' => string 'Sub1' (length=4)
      'link' => string 'menu4' (length=5)
      'show_condition' => string '1' (length=1)
      'parent' => string '2' (length=1)
      'class' => string '' (length=0)
      'data' => string '' (length=0)
      'des' => string '' (length=0)
  4 => 
    array (size=9)
      'id' => string '5' (length=1)
      'uid' => string '0' (length=1)
      'text' => string 'sub2' (length=4)
      'link' => string 'hhhhhhhhhhhh' (length=12)
      'show_condition' => string '1' (length=1)
      'parent' => string '2' (length=1)
      'class' => string '' (length=0)
      'data' => string '' (length=0)
      'des' => string '' (length=0)
  5 =>

请帮忙

Answer 1

你获得结果的方式太复杂了。我会在数据库中分离子项和父项菜单项，然后构建一个数组，以便视图可以更容易地解析它。试试这个：

DB：

enter image description here

然后你的模特：

function menus() {
    $this->db->select("*");
    $this->db->from("menu_parents");
    $q = $this->db->get();

    $final = array();
    if ($q->num_rows() > 0) {
        foreach ($q->result() as $row) {

            $this->db->select("*");
            $this->db->from("menu_children");
            $this->db->where("fk_parent_id", $row->parent_id);
            $q = $this->db->get();
            if ($q->num_rows() > 0) {
                $row->children = $q->result();
            }
            array_push($final, $row);
        }
    }
    return $final;
}

控制器：

public function get_menus() {
    $this->load->model('your_model');
    $menus = $this->your_model->menus();
    $data = array('menus' => $menus);
    $this->load->view('page1', $data);
}

查看：

<div class="navbar-collapse collapse">
    <ul class="nav navbar-nav">
        <?php foreach ($menus as $menu) { ?>
            <li><a><?= $menu->text ?></a></li>
            <?php
            if (isset($menu->children)) {
                foreach ($menu->children as $child) {
                    ?>
                    <li class="dropdown megamenu-fullwidth"> 
                        <a data-toggle="dropdown" class="dropdown-toggle" href="#"><</a>
                        <ul class="dropdown-menu"> 
                            <li class="megamenu-content">
                                <ul class="col-lg-2 col-sm-2 col-md-2">
                                    <li class="no-border"><p><strong>Head</strong></p></li>
                                    <li><a href="#">Descript<br></a></li>
                                </ul> 
                            </li> 
                        </ul>
                    </li>
                    <?php
                }
            }
            ?>
        <?php } ?>
    </ul>    
</div>

如何使用Codeigniter从数据库中创建动态导航菜单？

1 个答案: