Laravel在同一张桌子上回归父母的孩子

Ran*_*kin 10 php laravel eloquent blade laravel-5.1

使用Laravel 5.1,我试图从MySQL类别表创建一个菜单列表.我的服务提供者返回数据,但我不明白如何在foreach循环中创建子类别.当我执行循环时,只返回子查询的最后一行.任何指导将不胜感激.

类别表

id  | cat_name      | cat_parent_id
--- | --------------| ------------- 
1   | Parent Cat 1  | NULL 
2   | Parent Cat 2  | NULL 
3   | Child Cat 1   | 2 
4   | Child Cat 2   | 2 
5   | Parent Cat 3  | NULL 
6   | Child Cat 3   | 5
Run Code Online (Sandbox Code Playgroud)

期望的结果

Parent Cat 1
Parent Cat 2
    Child Cat 1
    Child Cat 2
Parent Cat 3
    Child Cat 3
Run Code Online (Sandbox Code Playgroud)

viewComposerServiceProvider.php

public function boot()
{
       $this->composeTopCategoryNavigation();
       $this->composeSubCategoryNavigation();
}

private function composeTopCategoryNavigation()
{
    view()->composer('partials.header', function($view)
        {
            $view->with('top_cats', Category::whereNull('cat_parent_id')->orderBy('cat_name', 'asc')->get());
        });
}

private function composeSubCategoryNavigation()
{
        view()->composer('partials.header', function($view)
        {
            $view->with('sub_cats', Category::whereNotNull('cat_parent_id')->orderBy('cat_name', 'asc')->get());
        });
}
Run Code Online (Sandbox Code Playgroud)

标题视图

<ul>
@foreach ($top_cats as $top_cat)
  <?php $top_cat_slug = str_slug( $top_cat->cat_name, "-"); ?>
  <li>{{ $top_cat->cat_name }}
    @foreach ($sub_cats as $sub_cat)
            @if ( $sub_cat->cat_parent_id === $top_cat->id )
                <ul>
                  <li{{ $sub_cat->cat_name }}</li>
                </ul>
                @endif
        @endforeach
   </li>
@endforeach
</ul>
Run Code Online (Sandbox Code Playgroud)

jed*_*ylo 22

首先,你所做的是效率低下的.您的视图会遍历每个父类别的所有子类别.如果您正确定义了关系并使用了Eloquent的急切加载,您可以更轻松地获取和访问子类别.

从定义关系开始:

class Category extends Model {
  //each category might have one parent
  public function parent() {
    return $this->belongsToOne(static::class, 'cat_parent_id');
  }

  //each category might have multiple children
  public function children() {
    return $this->hasMany(static::class, 'cat_parent_id')->orderBy('cat_name', 'asc');
  }
}
Run Code Online (Sandbox Code Playgroud)

一旦正确定义了关系,就可以像下面一样获取整个类别树:

view()->composer('partials.header', function($view) {
  $view->with('categories', Category::with('children')->whereNull('cat_parent_id')->orderBy('cat_name', 'asc')->get());
});
Run Code Online (Sandbox Code Playgroud)

由于父类别已包含子类,因此不需要第二个作曲家.

现在,您只需要在视图中显示类别:

<ul>
  @foreach ($categories as $parent)
    <li>{{ $parent->cat_name }}
      @if ($parent->children->count())
        <ul>
          @foreach ($parent->children as $child)
            <li>{{ $child->cat_name }}</li>
          @endforeach
        </ul>
      @endif
    </li>
  @endforeach
</ul>
Run Code Online (Sandbox Code Playgroud)