Mysql 以简单而雄辩的方式加入laravel 5.3

我想在我的左连接中添加一些额外的过滤器，但我不知道如何如此友好地帮助我。同时告诉我如何才能使这个查询有说服力。我的问题如下：

select * from `users`
join `halls` on `halls`.`user_id` = `users`.`id` 
left join `bookings` on `bookings`.`hall_id` = `halls`.`id` AND month(`bookings`.`date`) = 2 and day(`bookings`.`date`) = 4 and year(`bookings`.`date`) = 2017
join `user_role` on `user_role`.`user_id` = `users`.`id` 
join `roles` on `roles`.`id` = `user_role`.`role_id` 
where 
    `roles`.`id` = 2 AND

     (`bookings`.`id` is null OR `bookings`.`status` = 0 )

     group by users.id

用户和角色有多对多关系，用户和大厅一对多，大厅和预订也有一对多关系 用户模型关系

/**
     * Many-to-Many relations with Role.
     *
     * @return \Illuminate\Database\Eloquent\Relations\BelongsToMany
     */
public function roles(){
    return $this->belongsToMany(Role::class, 'user_role', 'user_id', 'role_id')->select('roles.name');
}
/**
 * One-to-Many relations with halls.
 *
 * @return \Illuminate\Database\Eloquent\Relations\hasMany
 */
public function halls(){
    return $this->hasMany(Hall::class);
}

霍尔模型关系

public function user(){

    return $this->belongsTo(User::class, 'user_id', 'id');
}
public  function  bookings(){
    return $this->hasMany(Booking::class);
}

预订模型房地产

public function hall(){
    return $this->belongsTo(Hall::class)->distinct();
}

---

我不知道为什么要使用没有任何聚合函数的group by。您的ORM如下所示

Users::join('halls', 'users.id', '=', 'halls.user_id')
leftJoin('bookings', function($join){
    $join->on('halls.id', '=', 'bookings.hall_id');
    $join->on(DB::raw('month(`bookings`.`date`) = 2 and day(`bookings`.`date`) = 4 and year(`bookings`.`date`) = 2017'));

})
->join('user_role', 'users.id', '=', 'user_role.user_id')
->join('roles', 'roles.id', '=', 'user_role.role_id')
->whereRaw('where 
    `roles`.`id` = 2 AND

     (`bookings`.`id` is null OR `bookings`.`status` = 0 )')->get();

---

你能粘贴你定义的模型和关系吗？