社交网络的Laravel雄辩ORM

Question

我的状态帖子有以下数据库设置。对于每一篇文章，用户可以喜欢这篇文章，评论这篇文章，甚至可以由作者在原始文章中添加标签。

我试图设置我的足智多谋的控制器后带回所有的数据通过JSON对象，但我不能正确地找到评论，喜欢或标记用户名。如果有区别的话，我会用哨兵2进行认证。

以下是数据库设置：

    CREATE TABLE Users (
        id INT NOT NULL AUTO_INCREMENT,
        first_name VARCHAR(30),
        last_name VARCHAR(30),
        many more...
    );

    CREATE TABLE Posts (
        postID INT NOT NULL AUTO_INCREMENT,
        caption VARCHAR(200), 
        description VARCHAR(200), 
        fromID INT(10) UNSIGNED NOT NULL,
        toID INT(10) UNSIGNED NOT NULL,
        icon VARCHAR(200),
        link VARCHAR(200),
        message TEXT,
        storyType INT,
        type ENUM ('LINK', 'PHOTO', 'STATUSUPDATE', 'VIDEO' ),
        createdTime DATE,
        PRIMARY KEY (postID),
        FOREIGN KEY (fromID) REFERENCES users (id),
        FOREIGN KEY (toID) REFERENCES users (id)
    ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

    CREATE TABLE Likes (
        likeID INT NOT NULL AUTO_INCREMENT,
        fromID INT(10) UNSIGNED NOT NULL,
        postID INT NOT NULL,
        createdDate DATE,
        PRIMARY KEY (likeID),
        FOREIGN KEY (fromID) REFERENCES users (id),
        FOREIGN KEY (postID) REFERENCES Posts (postID)
    ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

    CREATE TABLE Comments (
        commentID INT NOT NULL AUTO_INCREMENT,
        fromID INT(10) UNSIGNED NOT NULL,
        postID INT NOT NULL,
        comment TEXT,
        createdDate DATE,
        PRIMARY KEY (commentID),
        FOREIGN KEY (fromID) REFERENCES users (id),
        FOREIGN KEY (postID) REFERENCES Posts (postID)
    ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;


    CREATE TABLE Tags (
        tagID INT NOT NULL AUTO_INCREMENT,
        userID INT(10) UNSIGNED NOT NULL,
        postID INT NOT NULL,
        PRIMARY KEY (tagID),
        FOREIGN KEY (userID) REFERENCES users (id),
        FOREIGN KEY (postID) REFERENCES Posts (postID)
    ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

我的Post控制器，我只是有一个简单的页面，可以显示所有内容。我不想循环查看文件中的任何内容，我只想返回json完整对象。

class PostController extends BaseController {

    public function show($id)
    {
    $post = Post::with(array('comments', 'from', 'tags', 'likes'))->find($id);
    return View::make('samplepage')->with('data', $post);
    }
}

我的帖子模型：

class Post extends Eloquent {

    protected $table = 'Posts';
    protected $primaryKey = 'postID';

    public function comments() 
    {
       return $this->hasMany('Comment','postID');
     }

     public function tags() 
     {
       return $this->hasMany('Tag','postID');
     }

     public function likes() 
     {
       return $this->hasMany('Like','postID');
     }


    public function from() 
    {
        return $this->belongsTo('User', 'fromID')->select(array('id', 'first_name', 'last_name'));
    }

    public function users() 
    {
        return $this->belongsTo('User', 'fromID');
    }

}

注释模型：类注释扩展了雄辩的{

    protected $table = 'Comments';
    protected $primaryKey = 'commentID';

     public function post() {
       return $this->belongsTo('Post','fromID');
     }

     public function user() {
        return $this->belongsTo('User', 'fromID')->select(array('id', 'first_name', 'last_name'));
    }
}

标记模型：类标记扩展了雄辩的{

    protected $table = 'Tags';
    protected $primaryKey = 'tagID';

}

我甚至在我的用户模型中设置了以下内容，但没有区别。用户模型：

public function posts() {
   return $this->hasMany('Post','id');
 }

public function comments() {
 return $this->hasMany('Comment','id');
}

在这个设置中一切都很好，当我用下面的代码点击posts/2时，我得到了下面的对象$post=post:：with（数组（'comments'，'from'，'tags'，'likes'））-

{
   postID: "2",
   toID: "8",
   comments: [
      {
         commentID: "2",
         comment: "second comment",
         fromID: "1",
         postID: "2",
         createdDate: "2014-02-15"
      }
   ],
   from: {
      id: "4",
      first_name: Paul,
      last_name: Davis
   },
   tags: [
      {
         tagID: "1",
         userID: "2",
         postID: "2"
      },
      {
         tagID: "2",
         userID: "3",
         postID: "2"
      }
   ],
   likes: [
      {
         likeID: "1",
         fromID: "2",
         postID: "2",
         createdDate: "2013-01-04"
      },
      {
         likeID: "2",
         fromID: "3",
         postID: "2",
         createdDate: "2013-02-05"
      }
   ]
}

但是我想要的是以下内容，其中每个标记，like和注释连接第一个和最后一个名字，并将它们与对象一起返回。

{
   postID: "2",
   toID: "4",
   comments: [
      {
         commentID: "2",
         comment: "second comment",
         fromID: "1",
         from: {
              "name": "Jason Terry", 
              "id": "721286625"
            }, 
         postID: "2",
         createdDate: "2014-02-15"
      }
   ],
   from: {
      id: "4",
      first_name: Paul,
      last_name: Davis
   },
   tags: [
      {
         tagID: "1",
         userID: "2",
         from: {
              "name": "David Lee", 
              "id": "721286625"
            },
         postID: "2"
      },
      {
         tagID: "2",
         userID: "3",
         from: {
              "name": "Paul Pierce", 
              "id": "721286625"
            },
         postID: "2"
      }
   ],
   likes: [
      {
         likeID: "1",
         fromID: "2",
         from: {
              "name": "David Lee", 
              "id": "721286625"
            },
         postID: "2",
         createdDate: "2013-01-04"
      },
      {
         likeID: "2",
         fromID: "3",
         from: {
              "name": "Al Davis", 
              "id": "721286625"
            },
         postID: "2",
         createdDate: "2013-02-05"
      }
   ]
}

我已经搜索了Stackoverflow，无数的Laravel博客和官方文档两周了，我似乎无法解决这个问题。任何帮助都是非常感谢的。

更新：以下是Tony的回答

$post = Post::with(array('comments.users', 'from', 'tags.users', 'likes.users'))->find($id);  

然后我补充说

public function users() 
{ return $this->belongsTo('User', 'fromID')->select(array('id', 'first_name', 'last_name')); 
} 

到评论，标签，和喜欢的模型。这个物体现在工作得很好。

但是我的调试器显示了以下内容

select `id`, `first_name`, `last_name` from `users` where `users`.`id` in ('1')

select `id`, `first_name`, `last_name` from `users` where `users`.`id` in ('4')

select `id`, `first_name`, `last_name` from `users` where `users`.`id` in ('2', '3')

select `id`, `first_name`, `last_name` from `users` where `users`.`id` in ('2', '3')

简而言之，它在MyUsers表上运行4个查询。这不是多余的吗？它不应该对users表进行一次查询，而不是对原始post用户进行一次查询，对comments用户进行一次查询，对tags用户进行一次查询，对likes用户进行一次查询吗？

Answer 1

这是一个小小的老帖子，但你也可以这样做，

public function from() 
    {
        return $this->belongsTo('User', 'fromID')
                    ->select(array('id',DB::raw("CONCAT(firstname,' ', lastname) as name")));
    }

Answer 2

看起来您想要使用嵌套关系。

$post = Post::with('comments.from', 'from', 'tags.from', 'likes.from')->find($id);

您还需要将“from”关系编码到每个模型中。

获取连接的名称；在您的用户模型中需要以下内容

protected $appends = array('name');
protected $hidden = array('first_name', 'last_name');  //this is optional

public function getNameAttribute()
{
    return $this->attributes['first_name'] . ' ' . $this->attributes['last_name'];
}