Question

<ul>
      <li>Main Menu 1
        <ul>
          <li>Sub Menu 1.1
          </li>
          <li>Sub Menu 1.2
          </li>
        </ul>
      </li>
      <li>Main Menu 2
        <ul>
          <li>Sub Menu 2.1
          </li>
          <li>Sub Menu 2.2
          </li>
        </ul>
      </li>
    </ul>

上面是我想要制作的无序列表的表示。我能够使用foreach制作这种列表，一个用于获取主菜单的功能和一个用于获取子菜单的功能。我想征求关于如何制定某种条件以确定哪个子菜单适用于哪个菜单的建议，因为到目前为止所有子菜单同时显示所有菜单。特定子菜单应该有一个特定的菜单，这是我工作的代码

服务器端

function RetrieveAllMenu() {
  global $dbh;
  $stmt = $dbh - > prepare("SELECT * FROM userlist_tbl WHERE username = ?");
  $stmt - > bindValue(1, $_SESSION['login_user']);
  $stmt - > execute();
  $selected_row = $stmt - > fetch(PDO::FETCH_ASSOC);
  $mem_id = $selected_row['user_id'];
  if ($stmt - > execute()) {
    if ($stmt - > rowCount() > 0) {
      $stmt = $dbh - > prepare("SELECT * FROM rolemapping_tbl WHERE user_id = ?");
      $stmt - > bindValue(1, $mem_id);
      $stmt - > execute();
      $selected_row = $stmt - > fetch(PDO::FETCH_ASSOC);
      $rolelist_id = $selected_row['rolelist_id'];
      if ($stmt - > execute()) {
        $stmt = $dbh - > prepare("SELECT * FROM roledetails_tbl WHERE rolelist_id = ?");
        $stmt - > bindValue(1, $rolelist_id);
        $stmt - > execute();
        $selected_row = $stmt - > fetch(PDO::FETCH_ASSOC);
        if ($stmt - > execute()) {
          if ($stmt - > rowCount() > 0) {
            $menu_id = array();
            while ($selected_row = $stmt - > fetch(PDO::FETCH_ASSOC)) {
              $menu_id[] = array('menuid' => $selected_row['menulist_id'], );
            }
            $stmt = $dbh - > prepare("SELECT * FROM menulist_tbl WHERE menulist_id = :menuid");
            $menu_name = array();
            foreach($menu_id as $row) {
              $stmt - > execute(array(':menuid' => $row['menuid']));
              //while ($selected_row =$stmt->fetch(PDO::FETCH_COLUMN, 0)){
              while ($selected_row = $stmt - > fetch(PDO::FETCH_ASSOC)) {
                $menu_name[] = array('menuname' => $selected_row['menu_name'], 'menuurl' => $selected_row['menu_url'], 'menuflag' => $selected_row['menu_flag'], 'menuid' => $selected_row['menulist_id']);
              }
            }
            return $menu_name;
          }
        }
      }
    }
  }
}

function RetrieveAllSubMenu() {
  global $dbh;
  $menu_name = RetrieveAllMenu();
  $stmt = $dbh - > prepare("SELECT * FROM submenulist_tbl WHERE parent_id = :menuid");
  $submenu_name = array();
  foreach($menu_name as $row) {
      //$stmt->execute(array(':menuid' => $row['menuid']));
      $stmt - > bindValue(':menuid', $row['menuid'], PDO::PARAM_STR);
      $stmt - > execute();
      while ($selected_row = $stmt - > fetch(PDO::FETCH_ASSOC)) {
        $submenu_name[] = array('submenuname' => $selected_row['submenulist_name'], 'submenuurl' => $selected_row['submenulist_url'], 'submenuflag' => $selected_row['submenulist_flag']);
      }
    }
    //print_r($submenu_name);
  return $submenu_name;
  //return in_array($menuid, $submenu_name);
}

这是html端

<?php
echo'<ul>';   
foreach (RetrieveAllMenu() as $value){
//echo'<input  type="submit" value="'.$value['menuname'].'" name="'.$value['menuname'].'"/>';
echo'<li>';
echo '<a href="'.$value['menuurl'].'" id=""'.$value['menuname'].'"">'.$value['menuname'].'</a>';
echo'<ul>';
foreach (RetrieveAllSubMenu() as $value){
//if( $value['parentid'] === $value['menuid'] ){    
echo'<li><a href="'.$value['submenuurl'].'" id=""'.$value['submenuname'].'"">'.$value['submenuname'].'</a></li>';
//}
}
echo'</ul>';  
echo'</li>';         
}   
echo'</ul>';    
?>

这就是我的代码看起来应该如上所示。

<ul>
          <li>Main Menu 1
            <ul>
              <li>Sub Menu 1.1
              </li>
              <li>Sub Menu 1.2
              </li>
               <li>Sub Menu 2.1
              </li>
              <li>Sub Menu 2.2
              </li>
            </ul>
          </li>
          <li>Main Menu 2
            <ul>
              <li>Sub Menu 1.1
              </li>
              <li>Sub Menu 1.2
              </li>
              <li>Sub Menu 2.1
              </li>
              <li>Sub Menu 2.2
              </li>
            </ul>
          </li>
        </ul>

Answer 1

制作RetrieveAllSubMenu()功能，使其仅返回给定菜单的子菜单ID，而不是所有菜单

function RetrieveAllSubMenu($menu_id) {
  // your code
  $stmt = $dbh - > prepare("SELECT * FROM submenulist_tbl WHERE parent_id = :menuid");
      $stmt - > bindValue(':menuid', $menu_id, PDO::PARAM_STR);
 // more code
}

为您提供其背后的逻辑，但这将仅返回基于传递$menu_id的一个菜单的子菜单。

因此，在尝试获取子框架时，foreach应该看起来与此类似：

foreach (RetrieveAllSubMenu($value['menuid']) as $value){

我还建议您将$value的返回结果命名为：

foreach (RetrieveAllSubMenu($value['menuid']) as $submenu){

因为您还使用$value作为菜单和子菜单，如果您不完全理解内部foreach语句覆盖$ value时可能会导致问题

修改

您的foreach语句应类似于：

foreach (RetrieveAllMenu() as $menu){ echo'<li>'; echo '<a href="'.$menu['menuurl'].'" id=""'.$menu['menuname'].'"">'.$menu['menuname'].'</a>'; echo'<ul>'; foreach (RetrieveAllSubMenu($menu['menuid']) as $submenu){ echo'<li><a href="'.$submenu['submenuurl'].'" id=""'.$submenu['submenuname'].'"">'.$submenu['submenuname'].'</a></li>'; } }

在php pdo中使用嵌套的foreach嵌套列表

1 个答案: