使用模型类型根据其表获取另一个表

Question

我有一个Todo，而这个Todo 可以有一个SubTodo。每个SubTodo 可以是以下类型：

• 文字
• 下拉菜单
• 布尔值

这些SubTodo 中的每一个都有与其类型相关的细节，因此我决定将其分离到自己的表格中。数据库表结构如下：

todo
- id

sub_todo
- id
- todo_id
- type [dropdown, text, boolean]

sub_todo_dropdown
- id
- sub_todo_id

sub_todo_text
- id
- sub_todo_id

sub_todo_boolean
- id
- sub_todo_id

SubTodo 需要有一个名为 meta 的关系，它将解析特定于哪种类型的字段。我定义了以下关系：

模型/SubTodo.php

// The SubTodo will have meta information on other tables
// that will be depending on the type of the SubTodo.
// To fetch those meta information's we will
// have this set where will map the
// relationship to its model
const META_MODELS = [
    self::SUB_TODO_TYPE_TEXT => SubTodoText::class,
    self::SUB_TODO_TYPE_BOOLEAN => SubTodoBoolean::class,
    self::SUB_TODO_TYPE_DROPDOWN => SubTodoDropdown::class,
];

public function meta(): HasOne
{
    return $this->hasOne(self::META_MODELS[$this->type], 'sub_todo_id');
}

但是当我尝试加载这种关系时

$subTodos = SubTodo::with('meta')->paginate();

我收到"message": "Undefined index: "。做完之后

public function meta(): HasOne
{
    dd($this->type);
    return $this->hasOne(self::META_MODELS[$this->type], 'sub_todo_id');
}

我收到null。我最好的猜测是模型还没有加载，所以我需要先加载模型，然后调用meta：

$subTodos = SubTodo::limit(10)->get()->each(function (SubTodo $subtodo) {
   $subtodo->load('meta');
});

但是这种方法会导致N+1 问题。有什么方法可以实现加载meta 而无需先加载所有模型？这是one to one polymorphic relationship 的好用法吗？

Answer 1

好吧，我认为除了覆盖模型之外别无他法。

• 1 创建SubTodoBuilder 类：

  <?php
  
  
  namespace App\Builder;
  
  
  use Illuminate\Database\Eloquent\Builder;
  
  class SubTodoBuilder extends Builder
  {
    const RELATIONS = [
       "text" => "SubTodoText",
       "boolean" => "SubTodoBoolean",
       "dropdown" => "SubTodoDropdown"
    ];
  
   /**
    * Eager load the relationships for the models.
    *
    * @param array $models
    * @return array
    */
   public function eagerLoadRelations(array $models): array
   {
       foreach ($this->eagerLoad as $name => $constraints) {
           if ($name === "meta") {
               $groupedModels = collect($models)->groupBy("type");
               $models = [];
               foreach ($groupedModels as $type => $subModels) {
                   $relation = self::RELATIONS[$type];
                   $result = $this->eagerLoadRelation($subModels->all(), $relation, $constraints);
                   $models = array_merge($models, $result);
               }
           }
           //This part may need some modification
           if (strpos($name, '.') === false && $name !== "meta") {
               $models = $this->eagerLoadRelation($models, $name, $constraints);
           }
       }
       return $models;
   }
  }
  

• 2 在SubTodo 模型中：

  <?php
  
   namespace App\Models;
  
   use App\Builder\SubTodoBuilder;
   use Illuminate\Database\Eloquent\Builder;
   use Illuminate\Database\Eloquent\Factories\HasFactory;
   use Illuminate\Database\Eloquent\Model;
   use Illuminate\Database\Eloquent\Relations\HasOne;
  
   class SubTodo extends Model
   {
     use HasFactory;
  
  
     protected $fillable = [
         "todo_id",
         "type"
     ];
  
     public function SubTodoText(): HasOne
     {
         return $this->hasOne(SubTodoText::class);
     }
  
     public function SubTodoDropDown(): HasOne
     {
         return $this->hasOne(SubTodoDropdown::class);
     }
  
     public function SubTodoBoolean(): HasOne
     {
         return $this->hasOne(SubTodoBoolean::class);
     }
  
     public function meta(): HasOne
     {
         return $this->hasOne("meta");
     }
  
     /**
      * Create a new Eloquent query builder for the model.
      *
      * @param  \Illuminate\Database\Query\Builder  $query
      * @return Builder|static
      */
     public function newEloquentBuilder($query)
     {
         return new SubTodoBuilder($query);
     }
  

  }

• 3 在你的控制器中

  public function index()
  {
     $subTodos=SubTodo::with("meta")->get();
  }
  

现在，如果您有 30 个 SubTodo 需要获取，并且每个 10 个 SubTodo 具有三种类型之一，它将执行 1+3 个查询。