php - 4表Laravel 5.1之间的关系

我需要显示带有评论的feed，以及该feed上带有评论用户详细信息的total likes。
feeds表

| id | movie_id | user_id  | description |
|----|----------|----------|-------------|
| 1  | 1        | 1        | Lorem Ipsum |

注释表

| id | feed_id | user_id  | comment   |
|----|-------- |----------|-----------|
| 1  | 1       | 2        | comment 1 |
| 2  | 1       | 3        | comment 2 |

喜欢的桌子

| id | feed_id | user_id  |
|----|-------- |----------|
| 1  | 1       | 2        |
| 2  | 1       | 3        |

用户表

| id | username| email  |
|----|-------- |--------|
| 1  | a       | [email protected]|
| 2  | b       | [email protected]|
| 3  | c       | [email protected]|

关系
Feed.php

public function user () {
    return $this->belongsTo('App\User');
}

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

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

User.php

public function feeds () {
    return $this->belongsToMany('App\Feed');
}

public function like () {
    return $this->belongsTo('App\Like');
}

public function comment () {
    return $this->belongsTo('App\Comment');
}

Like.php

public function user () {
    return $this->belongsTo('App\User');
}

public function feed () {
    return $this->belongsTo('App\Feed');
}

Comment.php

public function user () {
    return $this->belongsTo('App\User');
}

public function feed () {
    return $this->belongsTo('App\Feed');
}

现在我需要获取所有feed（我已经完成了这项工作），其中包括comments count、likes count和评论过的用户详细信息。
我怎么能用雄辩的口才在一个查询中得到这一点。

最佳答案

试试这个

$commentsCount = \App\Models\Comment::select('feed_id',\DB::raw('count(id) as comments_count'))->groupBy('feed_id')->toSql();
    $likesCount = \App\Models\Like::select('feed_id',\DB::raw('count(id) as likes_count'))->groupBy('feed_id')->toSql();
    $records = \DB::table('feeds as f')
            ->leftJoin('comments as c','f.id','=','c.feed_id')
            ->leftJoin('users as u','c.user_id','=','u.id')
            ->leftJoin(\DB::raw('('.$commentsCount.') as k'),'f.id','=','k.feed_id')
            ->leftJoin(\DB::raw('('.$likesCount.') as l'),'f.id','=','l.feed_id')
            ->select('f.id as fid','f.description','u.id as uid','u.name','u.email','k.comments_count','l.likes_count')
            ->orderBy('fid')
            ->get();

    $transform = function(array $records){
        $records = collect($records)->groupBy('fid');
        return $records->transform(function($items){
            $feed['id'] = $items->first()->fid;
            $feed['description'] = $items->first()->description;
            $feed['count'] = [
                'likes' => is_null($items->first()->likes_count) ? 0 : $items->first()->likes_count,
                'comments' => is_null($items->first()->comments_count) ? 0 : $items->first()->comments_count,
            ];
            $feed['users'] = $items->transform(function($user){
                return is_null($user->uid) ? [] : ['id'=>$user->uid,'name'=>$user->name,'email'=>$user->email];
            });

            return $feed;
        });
    };

    return array_values($transform($records)->toArray());

您可以将闭包函数与其他函数交换。喜欢

$this->transform($records);

关于php - 4表Laravel 5.1之间的关系，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/34878105/