PHP/CodeIgniter关系型数据库数组搭建

Question

已解决!最终答案位于此问题的底部

我正在尝试使用 CodeIgniter 制作一个菜单生成器，出于某种原因，我似乎无法理解这个概念，尽管它看起来很简单(我是 PHP 和 CI 的新手)。这应该与 CodeIgniter 本身关系不大，因为我实际上只将它用于查询和 MVC 模式。

我有两个表:

• 菜单:

  • 编号
  • 姓名

• 菜单页面:

  • 编号
  • page_id(与 pages.id 相关)
  • menu_id(与 menus.id 相关)
  • item_name(它在菜单中的显示方式)
  • item_order(用于排序)
  • item_parent(用于在子菜单中嵌套项目)

编辑:这是我想要实现的结构:

array(  
    [0] => array(
        [menu_id]    => 1,
        [menu_name]  => 'Menu 1',
        [menu_pages] => array(
            [0] => array(
                [id]         => 1,
                [page_id]    => 1,
                [menu_id]    => 1,
                [item_name]  => 'Home',
                [item_order] => 0,
                [item_level] => 0,
                [parent_id]  => ''
            ),
            [1] => array(
                [id]         => 2,
                [page_id]    => 2,
                [menu_id]    => 1,
                [item_name]  => 'About',
                [item_order] => 0,
                [item_level] => 0,
                [parent_id]  => ''
            )
        )
    ),
    [1] => array(
        [menu_id]    => 2,
        [menu_name]  => 'Menu 2',
        [menu_pages] => array(
            [0] => array(
                [id]         => 3,
                [page_id]    => 3,
                [menu_id]    => 2,
                [item_name]  => 'Services',
                [item_order] => 0,
                [item_level] => 0,
                [parent_id]  => ''
            ),
            [1] => array(
                [id]         => 4,
                [page_id]    => 4,
                [menu_id]    => 2,
                [item_name]  => 'Contact',
                [item_order] => 0,
                [item_level] => 0,
                [parent_id]  => ''
            )
        )
    )
)

这是我目前所获得的(已更新):

function GetMenus() {
    $menus    = $this->db->get('menus');
    $menucols = $this->db->list_fields('menus');

    $pages    = $this->db->get('menu_pages');
    $pagecols = $this->db->list_fields('menu_pages');

    $arr      = array();
    $i        = 0;

    foreach($menus->result() as $menu) {
        foreach($pages->result() as $page) {
            foreach($pagecols as $col) {
                $arr[$i][$col] = $page->$col;
            }

            foreach($menucols as $cols) {
                $this->menus[$i][$cols] = $menu->$cols;

                if($arr[$i]['menu_id'] === $menu->id) {
                    $this->menus[$i]['menu_pages'] = $arr[$i];
                }
            }
        }

        $i++;
    }

    return $this->menus;
}

上面的实际输出是这样的:

Array(
    [0] => Array(
        [id] => 1,
        [name] => default,
        [menu_pages] => Array(
            Array( /* missing #1, showing #2/2 */
                [id] => 2
                [page_id] => 1
                [menu_id] => 1
                [item_name] => About
                [item_order] => 0
                [item_level] => 0
                [parent_id] => 
            )
        )
    ),
    [1] => Array(
        [id] => 2,
        [name] => menu2,
        [menu_pages] => Array(
            Array( /* missing #3, showing #4/4 */
                [id] => 4
                [page_id] => 3
                [menu_id] => 2
                [item_name] => Contact
                [item_order] => 0
                [item_level] => 0
                [parent_id] => 
            )
        )
    )
)

如您所见，这与我需要的非常接近，但是缺少一些项目，因为它们似乎在数组中被覆盖了(它只显示每个菜单的最后一个菜单项 - 似乎他们具有相同的 key )。

感谢所有帮助和建议!

编辑:这是最终的解决方案，大致基于 Mischa 的回答:

型号:

function GetMenus() {
    /* Yes, I know these can be chained, I unchained them to avoid
       horizontal scrolling on SO */
    $this->db->select('menus.name, menu_pages.*, pages.slug');
    $this->db->join('menu_pages', 'menu_pages.menu_id = menus.id');
    $this->db->join('pages', 'pages.id = menu_pages.page_id');
    $this->db->order_by('item_order', 'ASC');
    $menus  = $this->db->get('menus');
    $result = array();

    foreach($menus->result() as $menu) {
        $result[$menu->name][$menu->id] = array(
            'page_id'    => $menu->page_id,
            'menu_id'    => $menu->menu_id,
            'item_name'  => $menu->item_name,
            'item_slug'  => $menu->slug,
            'item_order' => $menu->item_order,
            'item_level' => $menu->item_level,
            'parent_id'  => $menu->parent_id
        );
    }

    return $result;
}

Controller :

$this->menu = $this->page->GetMenus();

查看:

<ul class="nav">
    <?php foreach($this->menu['default'] as $item) { ?>
        <li>
            <a href="<?php echo $item['item_slug']; ?>">
                <?php echo $item['item_name']; ?>
            </a>
        </li>
    <?php } ?>
</ul>

Answer 1

您有两个查询，但我想我会选择一个带有连接的查询。这使得代码更短、更简单。我没有测试下面的代码，但像这样的东西应该可以工作:

function GetMenus()
{
  $this->db->select('menus.name, menu_pages.*');
  $this->db->join('menu_pages', 'menu_pages.menu_id = menus.id');
  $this->db->order_by('menus.id');
  $q = $this->db->get('menus');

  $result = array();
  $current_menu_id = NULL;
  $i = -1;

  foreach($q->result() as $row)
  {
    if($current_menu_id !== $row->menu_id)
    {
      $i++;
      $result[] = array('menu_id' => $row->menu_id,
                        'menu_name' => $row->name,
                        'menu_pages' => array()
                  );
    }

    $result[$i]['menu_pages'][] = array('id' => $row->id,
                                        'page_id' => $row->page_id,
                                        'menu_id' => $row->menu_id,
                                        'item_name' => $row->item_name,
                                        'item_order' => $row->item_order,
                                        'item_level' => $row->item_level,
                                        'parent_id' => $row->parent_id
                                  );

    $current_menu_id = $row->menu_id;
  }

  return $result;
}