如何在拉拉维尔的口才中建立三张表之间的关系？

Question

我试图从一个与我目前使用的模型没有直接关系的表中检索数据。

我的数据结构：

表：posts

• id -整数
• 标题-字符串

表：post_stacks

• id -整数
• post_id -整数
• stack_id -整数

表：堆栈

• id -整数
• 体弦
• url -字符串

我的雄辩模型来自Post.php ( post表)，我正在尝试获取与post关联的所有堆栈(从堆栈表)。我只想在Post.php上声明我的关系，而不是我的枢轴表(post_stacks)。

我尝试过hasManyThrough，但是我的表的结构并不像Laravel所要求的那样“符合”需求，因为我需要一个外键在我的堆栈表上。

以下是我目前的实现：

Post.php

​

class Post extends Model
{

    protected $dates = [
      'created_at',
      'updated_at'
    ];

    public function user()
    {
        return $this->belongsTo(\App\User::class, 'user_id', 'id');
    }

    public function post_stacks()
    {
        return $this->hasMany(\App\PostStack::class);
    }

    public function post_os()
    {
        return $this->hasMany(\App\PostOS::class, 'post_id', 'id');
    }

    public function post_tags()
    {
        return $this->hasMany(\App\PostTag::class , 'post_id', 'id');
    }

    public function getCreatedAtAttribute($value)
    {
        return Carbon::parse($value)->toFormattedDateString();
    }

}

​

PostController.php

​

class PostController extends Controller
{
    public function index()
    {
        $posts = Post::all();

        foreach($posts as $post){
            $post->user;
            $post->created_at;
            $posts_os = $post->post_os;
            $post_stacks = $post->post_stacks;
            $post_tags = $post->post_tags;

            foreach($posts_os as $post_os){
                $os = OS::where('id', $post_os->os_id)->first();
                $post_os['body'] = $os['body'];
            }

            foreach($post_stacks as $post_stack){
                $stack = Stack::where('id', $post_stack->stack_id)->first();
                $post_stack['url'] = $stack['url'];
                $post_stack['body'] = $stack['body'];
            }

            foreach($post_tags as $post_tag){
                $tag = Tag::where('id', $post_tag->tag_id)->first();
                $post_tag['body'] = $tag['body'];
            }
        }

        return response()->json($posts);
    }

}

​

我的JSON数据响应

​

[  
   {  
      "id":1,
      "title":"Laravel + XAMPP",
      "user_id":1,
      "description":"I'll take you through the entire process of setting up a development environment for Laravel using XAMPP.",
      "created_at":"Jun 12, 2017",
      "updated_at":"2017-06-12 08:55:02",
      "user":{  
         "id":1,
         "name":"EpIEhg7ciO",
         "email":"AiyZXrubVG@gmail.com",
         "created_at":"2017-06-12 08:55:02",
         "updated_at":"2017-06-12 08:55:02"
      },
      "post_os":[  
         {  
            "id":1,
            "post_id":1,
            "os_id":1,
            "created_at":"2017-06-12 08:55:02",
            "updated_at":"2017-06-12 08:55:02",
            "body":"Windows"
         }
      ],
      "post_stacks":[  
         {  
            "id":1,
            "post_id":1,
            "stack_id":1,
            "created_at":"2017-06-12 08:55:02",
            "updated_at":"2017-06-12 08:55:02",
            "url":"laravel.svg",
            "body":"Laravel"
         },
         {  
            "id":2,
            "post_id":1,
            "stack_id":2,
            "created_at":"2017-06-12 08:55:02",
            "updated_at":"2017-06-12 08:55:02",
            "url":"xampp.svg",
            "body":"XAMPP"
         }
      ],
      "post_tags":[  
         {  
            "id":1,
            "post_id":1,
            "tag_id":1,
            "created_at":"2017-06-12 08:55:02",
            "updated_at":"2017-06-12 08:55:02",
            "body":"laravel"
         },
         {  
            "id":2,
            "post_id":1,
            "tag_id":2,
            "created_at":"2017-06-12 08:55:02",
            "updated_at":"2017-06-12 08:55:02",
            "body":"xampp"
         }
      ]
   },
   {  
      "id":2,
      "title":"Laravel + Vagrant",
      "user_id":1,
      "description":"I'll take you through the entire process of setting up a development environment for Laravel using Vagrant.",
      "created_at":"Jun 12, 2017",
      "updated_at":"2017-06-12 08:55:02",
      "user":{  
         "id":1,
         "name":"EpIEhg7ciO",
         "email":"AiyZXrubVG@gmail.com",
         "created_at":"2017-06-12 08:55:02",
         "updated_at":"2017-06-12 08:55:02"
      },
      "post_os":[  
         {  
            "id":2,
            "post_id":2,
            "os_id":1,
            "created_at":"2017-06-12 08:55:02",
            "updated_at":"2017-06-12 08:55:02",
            "body":"Windows"
         },
         {  
            "id":3,
            "post_id":2,
            "os_id":2,
            "created_at":"2017-06-12 08:55:02",
            "updated_at":"2017-06-12 08:55:02",
            "body":"Mac OS X"
         },
         {  
            "id":4,
            "post_id":2,
            "os_id":3,
            "created_at":"2017-06-12 08:55:02",
            "updated_at":"2017-06-12 08:55:02",
            "body":"Linux"
         }
      ],
      "post_stacks":[  
         {  
            "id":3,
            "post_id":2,
            "stack_id":1,
            "created_at":"2017-06-12 08:55:02",
            "updated_at":"2017-06-12 08:55:02",
            "url":"laravel.svg",
            "body":"Laravel"
         },
         {  
            "id":4,
            "post_id":2,
            "stack_id":3,
            "created_at":"2017-06-12 08:55:02",
            "updated_at":"2017-06-12 08:55:02",
            "url":"vagrant.png",
            "body":"Vagrant"
         }
      ],
      "post_tags":[  
         {  
            "id":3,
            "post_id":2,
            "tag_id":1,
            "created_at":"2017-06-12 08:55:02",
            "updated_at":"2017-06-12 08:55:02",
            "body":"laravel"
         },
         {  
            "id":4,
            "post_id":2,
            "tag_id":3,
            "created_at":"2017-06-12 08:55:02",
            "updated_at":"2017-06-12 08:55:02",
            "body":"vagrant"
         }
      ]
   }
]

​

我的JSON数据正是我想要的。我只是认为我的PostController实现效率低下，运行了太多的查询。我运行了太多的查询和嵌套循环。有没有一个干净的方法，我可以建立一个关系使用拉拉维尔的方法/关系？

谢谢!

Answer 1

你能在Stack模型上声明关系吗？您只排除在透视表上添加关系。如果是的话：

class Post extends Model
{
    public function stacks()
    {
        return $this->hasMany(\App\Stack::class);
    }
}

class Stack extends Model
{
    public function posts()
    {
        return $this->belongsToMany(\App\Post::class);
    }
}

根据PostStack模型中定义的字段，这应该是使关系正常工作所需的全部内容。此外，您还可以访问pivot属性，允许您：

$post->pivot->somePropertyOnStack

编辑

摘自这些文档，让你对雄辩如何决定关系有一个大致的认识：

记住，雄辩将自动确定正确的外键列的评论模式。按照惯例，雄辩者将取所属模型的“蛇案”名称，并以_id作为后缀。因此，对于这个示例，雄辩者将假设注释模型上的外键是post_id。

至于与联合表格的关系：

要确定关系的连接表的表名，雄辩者将按字母顺序加入两个相关的模型名称。

文档的这一部分将解释每种关系类型以及它们如何雄辩地发生：

https://laravel.com/docs/5.4/eloquent-relationships#defining-relationships