Question

我的状态帖子有以下数据库设置。对于每个帖子，用户可以喜欢帖子，评论帖子或甚至可以在作者的原始帖子中标记。

我正在尝试设置我的Resourceful控制器'Post'以通过JSON对象恢复所有数据但我无法正确找到评论，喜欢或标记用户名。如果有所作为，我正在使用Sentry 2进行身份验证。

这是数据库设置：

    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;

我的帖子控制器，我只是有一个简单的页面，吐出一切。我不想循环浏览视图文件中的任何内容，我只想恢复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');
    }

}

评论模型： class Comment扩展了Eloquent {

    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'));
    }
}

标签型号： class Tag扩展了Eloquent {

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

}

我甚至在我的用户模型中设置了以下内容，但它没有任何区别。用户模型：

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

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

这个设置一切都很好用，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（array（'comments'，'from'，'tags'，'likes'）） - ＆gt; find（$ id）; 返回View :: make（'samplepage'） - ＆gt; with（'data'，$ post）;

{
   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"
      }
   ]
}

但我想要的是以下内容，每个标签都有，并且注释连接名字和姓氏并将它们带回对象。

{
   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，这两周的官方文档，我似乎无法解决这个问题。任何帮助都非常感谢。

更新：在Tony的回答下面我添加了

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

然后我添加了

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

评论，标签和喜欢的模型。而且这个对象现在很有效。

但我的调试器显示以下内容

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，它在我的用户表上运行了4个查询。这不是多余的吗？不应该对用户表进行1次查询，而不是对原始帖子用户进行1次查询，对用户进行1次查询，对用户进行1次查询，对用户进行1次查询？

Answer 1

看起来你想使用嵌套关系。

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

您还需要将“from”关系编码到每个模型中。

获取连接名称;您需要在用户模型中使用以下内容

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'];
}

Answer 2

这是一个很老的帖子，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")));
    }

Laravel Eloquent社交网络ORM

2 个答案: