【发布时间】：2018-10-25 13:43:50
【问题描述】：
我想做一个 bootstrap 4 导航菜单，我有以下 SQL 查询，下面有一些代码，但我不知道该怎么做！
这些是表格
TABLE menu
--------------------------------------
| id | title | url |
| 1 | Home | index.php |
| 2 | Menu | # |
| 3 | Contact | # |
| 2 | Winkelwagen | winkelwagen.php |
--------------------------------------
TABLE categories
-------------------------------------
| id | title_cat | url | cparent_id |
| 1 | Auto's | # | 2 |
| 2 | Drank | # | 2 |
-------------------------------------
TABLE products
-------------------------------------
| id | product | url | pparent_id |
| 1 | Ferrari | # | 1 |
| 2 | Heineken | # | 2 |
-------------------------------------
这里是查询：
$query = "SELECT
X.level,
X.id,
X.name,
X.url,
X.parent_id
FROM
(
SELECT
1 AS LEVEL,
id AS id,
title AS NAME,
url AS url,
0 AS parent_id,
id AS id_1,
-1 AS id_2,
-1 AS id_3
FROM
menu
WHERE
1
UNION
SELECT
2 AS LEVEL,
id AS id,
title_cat AS NAME,
url AS url,
cparent_id AS parent_id,
cparent_id AS id_1,
id AS id_2,
-1 AS id_3
FROM
categories
WHERE
1
UNION
SELECT
3 AS LEVEL,
products.id AS id,
products.product AS NAME,
products.url AS url,
products.pparent_id AS parent_id,
categories.cparent_id AS id_1,
categories.id AS id_2,
products.id AS id_3
FROM
products
LEFT JOIN categories ON products.pparent_id = categories.id
WHERE
1
) X
WHERE
1
ORDER BY
id_1,
id_2,
id_3";
它给出了下表的级别（我也添加了 parent_id，但 parent_id buildTree($array) 进入循环）：
level id name url parent_id
1 1 Home index.php 0
1 2 Menu # 0
2 1 Auto's # 2
3 1 Ferrari # 1
2 2 Drank # 2
3 2 Heineken # 2
1 3 Contact contact.php 0
1 4 Winkelwagen winkelwagen.php 0
我希望导航菜单看起来像这样：
<li class="nav-item">
<a class="nav-link" href="index.php">Home</a>
</li>
<li class="nav-item dropdown">
<a class="nav-link dropdown-toggle" href="#" id="navbarDropdown" role="button" data-toggle="dropdown" aria-haspopup="true" aria-expanded="false">Menu</a>
<div class="dropdown-menu" aria-labelledby="navbarDropdown">
<div class="dropdown-submenu">
<a class="dropdown-item dropdown-toggle" href="#">Auto's</a>
<div class="dropdown-menu" aria-labelledby="navbarDropdown">
<a class="dropdown-item" href="#">Ferrari</a>
</div>
</div>
<div class="dropdown-submenu">
<a class="dropdown-item dropdown-toggle" href="#">Drank</a>
<div class="dropdown-menu" aria-labelledby="navbarDropdown">
<a class="dropdown-item" href="#">Heineken</a>
</div>
</div>
</div>
</li>
<li class="nav-item">
<a class="nav-link" href="#">Contact</a>
</li>
<li class="nav-item">
<a class="nav-link" href="winkelwagen.php">Winkelwagen</a>
</li>
我有以下代码，首先我们从上面已经看到的获取的查询中创建一个数组：
$sql = $pdo->prepare($query);
function menu_builder($sql) {
if ($sql->execute()) {
while ($row = $sql->fetch(PDO::FETCH_ASSOC)) {
$array[] = $row;
}
buildTree($array); // or menu_builder($sql);
}
}
下一个代码不起作用，因为它进入了一个无限循环（如果它起作用，我仍然需要使 html 正确:)：
function buildTree($array, $parent_id = 0, $parents = array()) {
if($parent_id == 0) {
foreach ($array as $element) {
if (($element['parent_id'] != 0) && !in_array($element['parent_id'], $parents)) {
$parents[] = $element['parent_id'];
}
}
}
$menu_html = '';
foreach($array as $element) {
if($element['parent_id'] == $parent_id) {
if(in_array($element['id'], $parents)) {
$menu_html .= '<li class="dropdown">';
$menu_html .= '<a href="'.$element['url'].'" class="dropdown-toggle" data-toggle="dropdown" role="button" aria-expanded="false">'.$element['name'].' <span class="caret"></span></a>';
}
else {
$menu_html .= '<li>';
$menu_html .= '<a href="' . $element['url'] . '">' . $element['name'] . '</a>';
}
if(in_array($element['id'], $parents)) {
$menu_html .= '<ul class="dropdown-menu" role="menu">';
$menu_html .= buildTree($array, $element['id'], $parents);
$menu_html .= '</ul>';
}
$menu_html .= '</li>';
}
}
return $menu_html;
}
这个是一个普通的<ul>/<li> 菜单，我不知道如何通过引导程序让它为我工作：
function menu_builder($sql) {
$level = 0;
if ($sql->execute()) {
while ($row = $sql->fetch(PDO::FETCH_ASSOC)) {
while($level < $row['level']) {
echo "<ul>" . PHP_EOL;
$level++;
}
while($level > $row['level']) {
echo "</ul>" . PHP_EOL;
$level--;
}
echo " <li>#" . $row['id'] . "->" . $row['name'] . "</li>" . PHP_EOL;
}
}
while($level-- > 0) {
echo "</ul>" . PHP_EOL;
}
}
如果您需要更多信息，请询问我，我尝试使用我正在尝试的代码和我正在使用的表格使问题尽可能清晰。
jQuery:
$('.dropdown-menu a.dropdown-toggle').on('click', function(e) {
if (!$(this).next().hasClass('show')) {
$(this).parents('.dropdown-menu').first().find('.show').removeClass("show");
}
var $subMenu = $(this).next(".dropdown-menu");
$subMenu.toggleClass('show');
$(this).parents('li.nav-item.dropdown.show').on('hidden.bs.dropdown', function(e) {
$('.dropdown-submenu .show').removeClass("show");
});
return false;
});
CSS：
.dropdown-submenu {
position: relative;
}
.dropdown-submenu a::after {
transform: rotate(-90deg);
position: absolute;
right: 6px;
top: .8em;
}
.dropdown-submenu .dropdown-menu {
top: 0;
left: 100%;
margin-left: .1rem;
margin-right: .1rem;
}
【问题讨论】：
标签： php css mysql bootstrap-4 navbar