Question

我被困在树状结构菜单中。

这是我的表

+----+----------------------+----------------------+-----------+
| id | title                | slug                 | parent_id |
+----+----------------------+----------------------+-----------+
|  1 | Cameras              | cameras              |      0     |
|  2 | Lighting             | lighting             |      0     |
|  3 | Portable Lights      | portable-lights      |      2     |
|  4 | Studio Lights        | studio-lights        |      2     |
|  5 | Lighting Accessories | lighting-accessories |      3     |
|  6 | Lens                 | lens                 |      0     |
|  7 | Tripods              | tripods              |      6     |
|  8 | Accessories          | accessories          |      7     |
|  9 | Miscellaneous        | miscellaneous        |      7     |
+----+----------------------+----------------------+-----------+

这是我的关系

public function parent(){
return $this->belongsTo('App\Category', 'parent_id');}
public function children(){
return $this->hasMany('App\Category', 'parent_id');}

通缉输出：

- Cameras
- Lighting
    -- Portable Lights
        --Lighting Accessories
    -- Studio Lights
- Lens
    -- Tripods
        - Accessories
        - Miscellaneous

我希望将上述结果放在包含（id和title）的数组中。

Answer 1

在模型中创建关系，如

 public function children() {
    return $this->hasMany('App\Category','parent_id','id') ;
 }

之后你的控制器动作获取类别结果

public function getCategoryList()
{
    $categories = Category::where('parent_id', '=', 0)->get();
    $allCategories = Category::pluck('title','id')->all();


    return view('categoryTreeMenu',compact('categories','allCategories'));
}

Answer 2

你应该尝试以下解决方案。

放置您的模型文件。

public function children() {
   return $this->hasMany('App\Category','parent_id','id') ;
}

您的控制器文件。

$parent = Task::where('parent_id', '=', 0)->get();
foreach($parent as $key => $value){
        if(isset($value->children)){
            $parent[$key]['children'] = $value->children;
            foreach ($parent[$key]['children'] as $key1 => $value1) {
                if(isset($value1->children)){
                }
            }
        }
}

您可以在 $ parent 数组中获取所有子项和子项数据。

Answer 3

好的，所以Collection管道非常适合您想要做的事情。到此为止;

// Retrieve all categories and then key them by their id
$categories = Category::all()->keyBy('id');
$categories->filter(function (Category $category) {
    // We want to only grab the categories that are children
    return $category->parent_id;
})->each(function (Category $category) use($categories) {
    // We not want to get the parent of this category
    $parent = $categories->get($category->parent_id);

    if ($parent) {
        // If the parent exists, and we haven't seen it yet, set a default collection
        if (! $parent->relationLoaded('children')) {
            $parent->setRelation('children', new Collection);
        }
        // We can be confident that it's a collection and won't auto load, so we add
        $parent->children->push($category);
    }
});

// Finally we want to return only root categories
return $categories->filter(function (Category $category) {
    return ! $category->parent_id;
});

我已经通过代码发表评论，希望能够解释发生了什么。我没有测试过这段代码，但它应该可以运行。

该集合应包含三个条目Cameras，Lighting和Lens。 Lighting->children应该包含2个条目，其中Portable Lights->children为1. Lens->children相同，依此类推。

树的结构菜单对象在laravel的列阵

3 个答案: