Question

我正在制作一个菜单系统，我想拥有无限的孩子。

我从数据库中读取并显示父母下的第二个孩子，但这只允许我有一个孩子。如果没有剩下的孩子，我该如何重复？

我可以复制上面的代码，但我想继续向数据库添加条目，并在需要时继续创建更多子代。

$get_top_menu = mysqli_query($con,"SELECT * FROM ss_folderstructure WHERE folderstructure_parent='0'");
while($found_top_menu = mysqli_fetch_array($get_top_menu))
        {

            echo "<li class='dropdown' ><a href=''><span class='icon-folder-open'></span>" . $found_top_menu['folderstructure_name'] . "</a>";
                echo "<ul>";

                        $get_mid_menu = mysqli_query($con,"SELECT * FROM ss_folderstructure WHERE folderstructure_parent='" . $found_top_menu['folderstructure_idno'] . "'");
                        while($found_mid_menu = mysqli_fetch_array($get_mid_menu))
                                    {       
                                    echo "<li><a href='bootstrap.html'>" . $found_mid_menu['folderstructure_name'] . "</a></li>";                                                                                       
                                    }
                echo "</ul>";
            echo "</li>";
        }

Answer 1

首先，您可以使用一个查询从数据库中获取所有数据。并从结果中形成数组。

像：

$result = array();
$get_top_menu = mysqli_query($con,"SELECT * FROM ss_folderstructure");
while($found_top_menu = mysqli_fetch_array($get_top_menu)) {
 $result[$found_top_menu['folderstructure_parent']][] = $found_top_menu;
}

在此之后你应该使用递归：

function doLoop($foo,$currId) {
   echo "<ul>";
   foreach ($foo[$currId] as $bar) {
    if (isset($bar['folderstructure_idno'])) {
      echo "<li>";
      doLoop($foo,$bar['folderstructure_idno']);
      echo "</li>";
    } else {
      echo "<li><a>{$bar['folderstructure_name']}</a></li>";
    }
   }
   echo "</ul>";
   return true;
}

doLoop($result,0);

从未结束菜单系统数据库驱动

1 个答案: