如何从数据库生成嵌套导航菜单

How to generate nested navigation menu from database

我将网站的页面存储在数据库中。使用 Laravel(尽管这对我的问题来说并不重要),我可以输出站点的层次结构并查看以下内容。 (如果需要,我可以灵活地更改数据库结构并添加更多列)

[
    {
        "ref_id": "1",
        "parent_id": "0",
        "name": "Item 1",
        "child_count": 0
    },
    {
        "ref_id": "2",
        "parent_id": "0",
        "name": "Item 2",
        "child_count": 2
    },
    {
        "ref_id": "3",
        "parent_id": "2",
        "name": "Item 2 Sub 1",
        "child_count": 0
    },
    {
        "ref_id": "4",
        "parent_id": "2",
        "name": "Item 2 Sub 2",
        "child_count": 0
    },
    {
        "ref_id": "5",
        "parent_id": "0",
        "name": "Item 3",
        "child_count": 2
    },
    {
        "ref_id": "6",
        "parent_id": "5",
        "name": "Item 3 Sub 1",
        "child_count": 0
    },
    {
        "ref_id": "7",
        "parent_id": "5",
        "name": "Item 3 Sub 2",
        "child_count": 0
    },
    {
        "ref_id": "12",
        "parent_id": "0",
        "name": "Item 4",
        "child_count": 0
    },
    {
        "ref_id": "13",
        "parent_id": "0",
        "name": "Item 5",
        "child_count": 3
    },
    {
        "ref_id": "14",
        "parent_id": "13",
        "name": "Item 5 Sub 1",
        "child_count": 0
    },
    {
        "ref_id": "15",
        "parent_id": "13",
        "name": "Item 5 Sub 2",
        "child_count": 2
    },
    {
        "ref_id": "16",
        "parent_id": "15",
        "name": "Item 5 Sub 2 SubSub 1",
        "child_count": 0
    },
    {
        "ref_id": "17",
        "parent_id": "15",
        "name": "Item 5 Sub 2 SubSub 2",
        "child_count": 0
    },
    {
        "ref_id": "18",
        "parent_id": "13",
        "name": "Item 5 Sub 3",
        "child_count": 0
    },
    {
        "ref_id": "19",
        "parent_id": "0",
        "name": "Item 6",
        "child_count": 0
    }
]

我需要遍历这些节点并输出如下所示的嵌套 UL 导航。 UL 可以嵌套多层,因此代码必须是动态的。

<ul>
    <li>Item 1</li>
    <li>Item 2
        <ul>
            <li>Item 2 Sub 1</li>
            <li>Item 2 Sub 2</li>
        </ul>
    </li> 
    <li>Item 3
        <ul>
            <li>Item 3 Sub 1</li>
            <li>Item 3 Sub 2</li>
        </ul>
    </li> 
    <li>Item 4</li>  
    <li>Item 5
        <ul>
            <li>Item 5 Sub 1</li>
            <li>Item 5 Sub 2
                <ul>
                    <li>Item 5 Sub 2 SubSub 1</li>
                    <li>Item 5 Sub 2 SubSub 2</li>
                </ul>
            </li>
            <li>Item 5 Sub 3</li>
        </ul>
    </li>   
    <li>Item 6</li>
</ul>  

到目前为止我所做的已经很接近了。这是目前最好的。这是 Laravel 的 blade 语法,但这是我所追求的逻辑,所以这也适用于原版 PHP。

<ul>
    @php
        $child_count = 0;
        $total_children = 0;
    @endphp
    @foreach($q_list as $row)
        @if($total_children)
            @php $child_count++; @endphp
        @endif
        <li>{{ $row->name }}
            @if($row->child_count)
                @php
                    $total_children = $row->child_count;
                    $child_count = 0;
                @endphp
                <ul>
            @elseif($total_children == $child_count && $total_children != 0)
                @php
                    $total_children = 0;
                    $child_count = 0;
                @endphp
                </ul>
            @endif
        @if($total_children == $child_count || $row->child_count == 0)
            </li>
        @endif
    @endforeach
</ul>

这导致:

<ul>
    <li>Item 1</li>
    <li>Item 2
        <ul>
            <li>Item 2 Sub 1</li>
            <li>Item 2 Sub 2
        </ul>
    </li>
    <li>Item 3
        <ul>
            <li>Item 3 Sub 1</li>
            <li>Item 3 Sub 2
        </ul>
    </li>
    <li>Item 4</li>
    <li>Item 5
        <ul>
            <li>Item 5 Sub 1</li>
            <li>Item 5 Sub 2
                <ul>
                    <li>Item 5 Sub 2 SubSub 1</li>
                    <li>Item 5 Sub 2 SubSub 2
                </ul>
            </li>
            <li>Item 5 Sub 3</li>
            <li>Item 6</li>
</ol>

我觉得脚本很乱,但更重要的是,这里有两个问题。首先,每个嵌套 UL 的最后一项都省略了结束 </li>。其次,每个嵌套的 <ul> 只关闭一次。如果嵌套的 <ul> 仅在第二级,但任何更高级别(如 Item 5 Sub 2 SubSub *)没有足够的关闭 <li></ul> 标签导致 Item 6 处于错误级别,那很好. (即仍然是项目 5 的子项)

任何人都可以帮助填补空白或让我知道实现此目标的更好方法。谢谢

你可以实现递归逻辑。

创建助手

function generate_tree($categories)
{
    foreach ($categories as $category) {
        echo '<li id="categoryId_' . $category->id . '">';
        echo $category->name;
        if ($category->children) {
            echo '<ul>';
            generate_tree($category->children);
            echo '</ul>';
        }
        echo '</li>';
    }
}

正在观看

<ul>
   {!! generate_tree($categories) !!}
</ul>