使用php和mysql生成一個簡單的第二級下拉菜單

Question

我有一個top_menu和一個名為table的菜單。 我想通過菜單id-s獲得一個簡單的下拉菜單。 這段代碼的問題是，它回顯了3個“Információk”菜單項，並把almenu表中的3個菜單項放入其中。

該代碼現在可以執行以下操作：（它從top_menu表中彈出了3倍的菜單項，作為導航欄中的新菜單項）。

不是：Informáciok（Alapadatok，Kollégium，Osztályok）（這就是我想要的）

<?php
                $menu_sql = 
                "
                    SELECT 

                        top_menu.top_menu_nev,
                        top_menu.top_menu_seo,
                        top_menu.top_menu_dropdown,
                        almenu.almenu_nev,
                        almenu.almenu_seo

                    FROM top_menu LEFT JOIN almenu ON top_menu.top_menu_id = almenu.almenu_parent

                    WHERE top_menu.menu_status = 1 AND top_menu.menu_position = 1

                    ORDER BY top_menu.top_menu_sorrend ASC

                ";
                $get_menu = mysqli_query($kapcs, $menu_sql) or die(mysqli_error($kapcs));
                while($top_menu = mysqli_fetch_assoc($get_menu))
                {
                    if($top_menu['top_menu_dropdown'] == 0 )
                    {
                        echo '<li><a href="'.$host.'/'.$top_menu['top_menu_seo'].'" title="'.$top_menu['top_menu_nev'].'" class="top_menu_to_link">'.$top_menu['top_menu_nev'].'</a></li>';
                    }
                    else
                    {
                        echo '<li class="dropdown">';                       
                            echo '<a href="#" title="" class="top_menu_to_link dropdown-toggle" data-toggle="dropdown" role="button" aria-haspopup="true" aria-expanded="false">'.$top_menu['top_menu_nev'].'</a>'; 
                            echo '<ul class="dropdown-menu">';
                                echo '<li><a href="'.$host.'/'.$top_menu['almenu_seo'].'" title="'.$top_menu['almenu_nev'].'" class="top_menu_to_link">'.$top_menu['almenu_nev'].'</a></li>';
                            echo '</ul>';                   
                        echo '</li>';
                    }
                }
                ?>

Answer 1

問題是您一次又一次地遍歷整個導航。

以下代碼可讓您生成新的li項目：

            echo '<li class="dropdown">';                       
            echo '<a href="#" title="" class="top_menu_to_link dropdown-toggle" data-toggle="dropdown" role="button" aria-haspopup="true" aria-expanded="false">'.$top_menu['top_menu_nev'].'</a>'; 
            echo '<ul class="dropdown-menu">';

            $get_menu = mysqli_query($kapcs, $menu_sql) or die(mysqli_error($kapcs));
            while($top_menu = mysqli_fetch_assoc($get_menu))
            {
                if($top_menu['top_menu_dropdown'] == 0 )
                {
                    echo '<li><a href="'.$host.'/'.$top_menu['top_menu_seo'].'" title="'.$top_menu['top_menu_nev'].'" class="top_menu_to_link">'.$top_menu['top_menu_nev'].'</a></li>';
                }
                else
                {
                            echo '<li><a href="'.$host.'/'.$top_menu['almenu_seo'].'" title="'.$top_menu['almenu_nev'].'" class="top_menu_to_link">'.$top_menu['almenu_nev'].'</a></li>';

                }
            }
            echo '</ul>';
            echo '</li>';