根据laravel查询中的用户id获取推荐用户

class User extends Model
{
    public function followers()
    {
        return $this->belongsToMany(User::class, 'follows', 'followee_id', 'follower_id');
    }

    public function followees()
    {
        return $this->belongsToMany(User::class, 'follows', 'follower_id', 'followee_id');
    }
}

use App\Models\User;

function getRecommendedUsers($userId)
{
    $user = User::find($userId);

    if (!$user) {
        return [];
    }

    // 获取当前用户的关注者
    $followers = $user->followers;

    // 获取当前用户关注的人
    $followees = $user->followees;

    // 基于共同关注者推荐用户
    $recommendedUsers = User::whereNotIn('id', [$userId])
        ->whereIn('id', function ($query) use ($followers) {
            $query->select('followee_id')
                ->from('follows')
                ->whereIn('follower_id', $followers->pluck('id')->toArray())
                ->groupBy('followee_id')
                ->havingRaw('COUNT(DISTINCT follower_id) > 1');
        })
        ->get();

    return $recommendedUsers;
}