Laravel - 模型自引用belongsToMany与其他关系和具有多个限制的查询

Question

我有 3 张桌子：

建筑物：

+-----+-------------+-----------+--------------------------+------------+---------------+----------------------+---------------------+
| id  | order_flag  |   type    |           key            | unit_cost  | unit_produce  |     created_at       |     updated_at      |
+-----+-------------+-----------+--------------------------+------------+---------------+----------------------+---------------------+
|  1  |         10  | resource  | building1                |        10  |            0  | 2015-01-30 08:54:44  | 2015-01-30 08:54:44 |
|  2  |         20  | resource  | building2                |         5  |            0  | 2015-01-30 08:54:44  | 2015-01-30 08:54:44 |
|  3  |         30  | resource  | building3                |        10  |            0  | 2015-01-30 08:54:44  | 2015-01-30 08:54:44 |
|  4  |         10  | basic     | building4                |         0  |           25  | 2015-01-30 08:54:44  | 2015-01-30 08:54:44 |
+-----+-------------+-----------+--------------------------+------------+---------------+----------------------+---------------------+

建筑用户：

+-----+--------------+----------+--------+----------------------+---------------------+
| id  | building_id  | user_id  | level  |     created_at       |     updated_at      |
+-----+--------------+----------+--------+----------------------+---------------------+
|  1  |           2  |      12  |     3  | 2015-01-30 08:52:57  | 2015-01-30 08:55:37 |
|  2  |           4  |      12  |     1  | 2015-01-30 08:53:53  | 2015-01-30 08:53:53 |
|  3  |           1  |      12  |     2  | 2015-01-30 08:54:08  | 2015-01-30 08:55:10 |
+-----+--------------+----------+--------+----------------------+---------------------+

建筑要求：

+-----+--------------+-------------+--------+----------------------+---------------------+
| id  | building_id  | require_id  | level  |     created_at       |     updated_at      |
+-----+--------------+-------------+--------+----------------------+---------------------+
|  1  |           1  |          4  |     1  | 2015-01-30 08:54:44  | 2015-01-30 08:54:44 |
|  2  |           3  |          1  |     5  | 2015-01-30 08:54:44  | 2015-01-30 08:54:44 |
+-----+--------------+-------------+--------+----------------------+---------------------+

我在建筑模型中写下这个关系：

    # Building Require another Building(s)
    public function requires() {
        return $this->belongsToMany('Building', 'building_require', 'building_id', 'require_id')->withPivot(array(
            'level',
            'updated_at',
            'created_at'
        ));
    }

    public function users() {
        return $this->belongsToMany('User')->withPivot(array('level'));
    }

现在我想从桌子建筑物中选择所有建筑物...... 以下限制：

1. 如果 building_require 表中没有来自该建筑物的条目，则该建筑物可以在结果中。

2. 如果该建筑物在 building_require 表中有条目，则必须检查 building_user 表中是否有条目，其中 building_id 与 building_required 表中的 required_id 相同。

（例如 building_id 1 需要 building_id 4，而 building_id 3 需要 building_id 1 的条目。）

除此之外（这是困难的部分），必须检查 building_user 表中的条目是否具有 level 值 >= 作为 building_require 表中定义的级别。

总结一下：

在 这种情况 building_id 1 需要在 building_user 中的条目 building_id 为 4 并且级别至少为 1。 building_id 3 需要一个条目 building_user 其中 building_id 为 1 且级别至少为 5。因此 - building_id 1 和 3 具有所需的建筑物。建筑与 id 2 和 4 在结果中没有限制。 在这种情况下，只有 building_id 3 是不允许的。因为 building_user 表中 building_id 1 的级别只有 2 而不是至少 5。

对我来说这很困难，我认为这不是最好的解决方案，遗憾的是没有检查级别值：

# all buildings there has a required building:
$allRequiredBuildingIds = DB::table('building_require')->lists('building_id');

# if user has buildings, this buildings must remove from $allRequiredBuildingIds
if(Sentry::getUser()->buildings()->count() > 0) {
    # list all building_id 's of the user_building table
    $userBuildingIds = Sentry::getUser()->buildings()->lists('building_id');
    # list all building_id's of the buildings there are required and already stored in the building_user table
    $userBuildingRequiredIds = DB::table('building_require')->whereIn('require_id', $userBuildingIds)->lists('building_id');
    # there are all the building_id's which are not allowed to display
    $allRequiredBuildingIds = array_values(array_diff($allRequiredBuildingIds,$userBuildingRequiredIds));
}

# if there are buildings there are not allowed to display?
if(count($allRequiredBuildingIds) > 0) {
    $buildings = Building::whereType($type)->whereNotIn('id', $allRequiredBuildingIds)->paginate(10);
}else{
    $buildings = Building::whereType($type)->paginate(10);
}

有人可以帮我优化它并集成级别检查吗？ 我的头在怦怦直跳。

Answer 1

我有一个理论认为这可能有效，但我尚未对其进行测试。我认为关键是只在 whereHas 闭包中使用 Join ，从而让 Eloquent 构建器保持活跃。

Building::doesntHave('requires')
    ->orWhereHas('users', function($q) {
        $q->join('building_require', 'building_user.building_id', '=', 'building_require.required_id')
          ->where('building_user.level', '>=', 'building_require.level');
    })->get();

它会生成这样的查询：

select * from `buildings` where (
  select count(*) from `building_require` as `x` where `x`.`building_id` = `buildings`.`id`
) < 1 or (
  select count(*) from `users` 
  inner join `building_user` on `users`.`id` = `building_user`.`user_id` 
  inner join `building_require` on `building_user`.`building_id` = `building_require`.`required_id` 
  where `building_user`.`building_id` = `buildings`.`id` 
  and `building_user`.`level` >= building_require.level
) >= 1

---

不幸的是，这并不能解决问题。它将返回 building_require.required_id 建筑物而不是 building_require.building_id 建筑物。我设法在原始 SQL 中找出正确的查询，但我不确定如何在 Eloquent 中修复它：

SQLFiddle

select * from `buildings` where (
  select count(*) from `building_require` as `x` where `x`.`building_id` = `buildings`.`id`
) < 1 or (
  select count(*) from `users` 
  inner join `building_user` on `users`.`id` = `building_user`.`user_id` 
  inner join `building_require` on `building_user`.`building_id` = `building_require`.`required_id` 
  where building_require.building_id = buildings.id
  and `building_user`.`level` >= building_require.level
) >= 1;