【发布时间】:2021-12-12 12:08:18
【问题描述】:
我有一个嵌套的下拉菜单,它显示来自数据库的数据,我想在选择它时获取数据的 ID,我也希望能够多选,但我不能,因为我什至无法获得对应于该特定数据的 id。
我正在尝试实现一种用户可以轻松浏览数据并能够进行多选的方式,到目前为止我已经实现了浏览。
这是模型
`
<?php
namespace App\Models;
use Illuminate\Database\Eloquent\Factories\HasFactory;
use Illuminate\Database\Eloquent\Model;
class Menu extends Model
{
use HasFactory;
protected $fillable = ['functional_area','parent_id'];
public function parent()
{
return $this->hasOne('App\Models\Menu', 'id', 'parent_id')->orderBy('sort_order');
}
public function children()
{
return $this->hasMany('App\Models\Menu', 'parent_id', 'id')->orderBy('sort_order');
}
public static function tree()
{
return static::with(implode('.', array_fill(0, 100, 'children')))->where('parent_id', '=', '0')->orderBy('sort_order')->get();
}
}`
这是我的控制器 `
<?php
namespace App\Http\Controllers;
use Illuminate\Http\Request;
class MenuController extends Controller
{
public function getMenu()
{
$menu = new \App\Models\Menu;
$menuList = $menu->tree();
return view('index')->with('menulist', $menuList);
}
}
` 这是它的索引
`
<body>
<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.5.2/css/bootstrap.min.css" />
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/font-awesome/5.14.0/css/all.min.css" integrity="sha512-1PKOgIY59xJ8Co8+NE6FZ+LOAZKjy+KY8iq0G4B3CyeY6wYHN3yt9PW0XpSriVlkMXe40PTKnXrLnZ9+fkDaog==" crossorigin="anonymous" />
<style>
.dropdown-menu .nav-item a { color: #000 !important; }
.dropdown-toggle:after { content: none; }
.dropdown-menu .dropdown-menu { margin-left: 0; margin-right: 0; }
.dropdown-menu li { position: relative }
.nav-item .submenu { display: none; position: absolute; left: 100%; top: -7px; }
.dropdown-menu>li:hover { background-color: #f1f1f1; }
.dropdown-menu>li:hover>.submenu { display: block; }
</style>
<nav class="navbar navbar-expand-lg navbar-dark bg-dark">
<a class="navbar-brand"></a>
<ul class="navbar-nav mr-auto">
@each('submenu', $menulist, 'menu', 'empty')
</ul>
</nav>
<script src="https://code.jquery.com/jquery-3.5.1.min.js" integrity="sha256-9/aliU8dGd2tb6OSsuzixeV4y/faTqgFtohetphbbj0=" crossorigin="anonymous"></script>
<script src="https://stackpath.bootstrapcdn.com/bootstrap/4.5.2/js/bootstrap.min.js"></script>
<script type="text/javascript">
$(document).on('click', '.dropdown-menu', ($event) => $event.stopPropagation());
if ($(window).width() < 992) {
$('.dropdown-menu a').click(($event) => {
$event.preventDefault();
if ($(this).next('.submenu').length) {
$(this).next('.submenu').toggle();
}
$('.dropdown').on('hide.bs.dropdown', () => $(this).find('.submenu').hide());
});
}
</script>
</body>
这是一个子菜单
@if ((count($menu->children) > 0) AND ($menu->parent_id > 0))
<li class="nav-item dropdown">
<a href="" class="nav-link dropdown-toggle" role="button" data-toggle="dropdown">
{{ $menu->functional_area }}
@if(count($menu->children) > 0)
<i class="fa fa-caret-right"></i>
@endif
</a>
@else
<li class="nav-item @if($menu->parent_id === 0 && count($menu->children) > 0) dropdown @endif">
<a href="" class="nav-link dropdown-toggle" data-toggle="dropdown">
{{ $menu->functional_area }}
@if(count($menu->children) > 0)
<i class="fa fa-caret-down"></i>
@endif
</a>
@endif
@if (count($menu->children) > 0)
<ul class="@if($menu->parent_id !== 0 && (count($menu->children) > 0)) submenu @endif dropdown-menu" aria-labelledby="dropdownBtn">
@foreach($menu->children as $menu)
@include('submenu', $menu)
@endforeach
</ul>
@endif
</li>
希望有人能帮帮我。
【问题讨论】:
-
您可以使用带有 parent_id 列(可为空)的单个表。如果 parent_id 为 null,则它是父菜单项。如果有 parent_id 它是一个子项。
-
已经完成了,我的问题是现在如何保存。
标签: php laravel bootstrap-4 laravel-8