我的数据库表格如下:
对象表(Object Table):
+----+---------+
| id | title |
+----+---------+
| 1 | Apple |
| 3 | Carrot |
| 4 | Dill |
| 5 | Egg |
| 6 | Fred |
| 7 | Goat |
| 8 | Harry |
| 9 | Igloo |
| 10 | Jason |
| 11 | Klaus |
| 12 | Banana |
| 15 | Oyster1 |
| 16 | Oyster2 |
+----+---------+
对象链接表:
+----+---------+--------------+
| id | obj_id | obj_link_id |
+----+---------+--------------+
| 1 | 1 | 12 |
| 2 | 1 | 5 |
| 3 | 3 | 1 |
| 4 | 3 | 12 |
| 5 | 3 | 3 |
| 6 | 4 | 1 |
| 7 | 4 | 5 |
| 8 | 5 | 6 |
| 9 | 6 | 7 |
| 10 | 7 | 7 |
| 11 | 7 | 8 |
| 12 | 9 | 12 |
| 13 | 9 | 5 |
| 14 | 10 | 1 |
| 15 | 10 | 5 |
| 16 | 10 | 8 |
| 17 | 11 | 1 |
| 18 | 11 | 5 |
| 19 | 11 | 10 |
| 20 | 12 | 3 |
| 21 | 15 | 16 |
| 22 | 16 | 15 |
+----+---------+--------------+
那么,从表格中可以看到对象1与对象12和5都有链接。
我的MySQL查询代码如下:
select object.id, title, obj_link_id
from object
left join object_links ON object.id = object_links.object_id
order by object.id
生成以下表格:
+----+---------+--------------+
| id | title | obj_link_id |
+----+---------+--------------+
| 1 | Apple | 12 |
| 1 | Apple | 5 |
| 3 | Carrot | 1 |
| 3 | Carrot | 12 |
| 3 | Carrot | 3 |
| 4 | Dill | 1 |
| 4 | Dill | 5 |
| 5 | Egg | 6 |
| 6 | Fred | 7 |
| 7 | Goat | 7 |
| 7 | Goat | 8 |
| 8 | Harry | NULL |
| 9 | Igloo | 12 |
| 9 | Igloo | 5 |
| 10 | Jason | 1 |
| 10 | Jason | 5 |
| 10 | Jason | 8 |
| 11 | Klaus | 1 |
| 11 | Klaus | 5 |
| 11 | Klaus | 10 |
| 12 | Banana | 3 |
| 15 | Oyster1 | 16 |
| 16 | Oyster2 | 15 |
+----+---------+--------------+
我在 PHP 中使用:
$objects = $stmt->fetchAll(PDO::FETCH_CLASS);
我不确定是否有更好的方法来获取这些内容,所以非常欢迎提出建议。
print_r($objects)
的输出结果为:
Array
(
[0] => stdClass Object
(
[id] => 1
[title] => Apple
[obj_link_id] => 12
)
[1] => stdClass Object
(
[id] => 1
[title] => Apple
[obj_link_id] => 5
)
[2] => stdClass Object
(
[id] => 3
[title] => Carrot
[obj_link_id] => 1
)
[3] => stdClass Object
(
[id] => 3
[title] => Carrot
[obj_link_id] => 12
)
[4] => stdClass Object
(
[id] => 3
[title] => Carrot
[obj_link_id] => 3
)
[5] => stdClass Object
(
[id] => 4
[title] => Dill
[obj_link_id] => 1
)
[6] => stdClass Object
(
[id] => 4
[title] => Dill
[obj_link_id] => 5
)
[7] => stdClass Object
(
[id] => 5
[title] => Egg
[obj_link_id] => 6
)
[8] => stdClass Object
(
[id] => 6
[title] => Fred
[obj_link_id] => 7
)
[9] => stdClass Object
(
[id] => 7
[title] => Goat
[obj_link_id] => 7
)
[10] => stdClass Object
(
[id] => 7
[title] => Goat
[obj_link_id] => 8
)
[11] => stdClass Object
(
[id] => 8
[title] => Harry
[obj_link_id] =>
)
[12] => stdClass Object
(
[id] => 9
[title] => Igloo
[obj_link_id] => 12
)
[13] => stdClass Object
(
[id] => 9
[title] => Igloo
[obj_link_id] => 5
)
[14] => stdClass Object
(
[id] => 10
[title] => Jason
[obj_link_id] => 1
)
[15] => stdClass Object
(
[id] => 10
[title] => Jason
[obj_link_id] => 5
)
[16] => stdClass Object
(
[id] => 10
[title] => Jason
[obj_link_id] => 8
)
[17] => stdClass Object
(
[id] => 11
[title] => Klaus
[obj_link_id] => 1
)
[18] => stdClass Object
(
[id] => 11
[title] => Klaus
[obj_link_id] => 5
)
[19] => stdClass Object
(
[id] => 11
[title] => Klaus
[obj_link_id] => 10
)
[20] => stdClass Object
(
[id] => 12
[title] => Banana
[obj_link_id] => 3
)
[21] => stdClass Object
(
[id] => 15
[title] => Oyster1
[obj_link_id] => 16
)
[22] => stdClass Object
(
[id] => 16
[title] => Oyster2
[obj_link_id] => 15
)
)
请注意,括号中的数字仅为数组索引,而非对象 ID 编号,请不要被索引所迷惑。我正在尝试找到一种确定哪些是已链接对象和未链接对象的方法。根据上述情况,对象应该分别如下:
**Linked:**
Apple
Banana
Carrot
Egg
Fred
Goat
Harry
**Not Linked:**
Dill
Igloo
Jason
Klaus
Oyster1
Oyster2
我主要的问题是:
在 PHP 中,当每个对象都可能有多个链接时,如何创建一个循环以遍历这样的结构?最终,我想要产生两个对象集合,一个包含已链接的对象,另一个包含未链接的对象。示例集合可能如下所示:
stdClass Object
(
[LinkedElements] => stdClass Object
(
[1] => stdClass Object
(
[id] => 1
[name] => Apple
[link] => Array
(
[0] => 14
[1] => 5
)
)
[14] => stdClass Object
(
[id] => 14
[name] => Banana
[link] => Array
(
[0] => 3
)
)
[3] => stdClass Object
(
[id] => 3
[name] => Carrot
[link] => Array
(
[0] => 1
[1] => 14
[2] => 3
)
)
[5] => stdClass Object
(
[id] => 5
[name] => Egg
[link] => Array
(
[0] => 6
)
)
[6] => stdClass Object
(
[id] => 6
[name] => Fred
[link] => Array
(
[0] => 7
)
)
[7] => stdClass Object
(
[id] => 7
[name] => Goat
[link] => Array
(
[0] => 7
[1] => 8
)
)
[8] => stdClass Object
(
[id] => 8
[name] => Harry
)
)
[UnLinkedElements] => stdClass Object
(
[4] => stdClass Object
(
[id] => 4
[name] => Dill
[link] => Array
(
[0] => 1
[1] => 5
)
)
[9] => stdClass Object
(
[id] => 9
[name] => Igloo
[link] => Array
(
[0] => 14
[1] => 5
)
)
[10] => stdClass Object
(
[id] => 10
[name] => Jason
[link] => Array
(
[0] => 1
[1] => 5
[2] => 8
)
)
[11] => stdClass Object
(
[id] => 11
[name] => Klaus
[link] => Array
(
[0] => 1
[1] => 5
[2] => 10
)
)
[15] => stdClass Object
(
[id] => 15
[name] => Oyster1
[link] => Array
(
[0] => 16
)
)
[16] => stdClass Object
(
[id] => 16
[name] => Oyster2
[link] => Array
(
[0] => 15
)
)
)
)
请注意:
- 导航是从对象到链接,而不是相反的方向。
- 让一个对象指向自身是可以的(就像对象7一样)。
- 上面的示例结构(在我的主要问题下面)仅供参考,我也欢迎其他建议。
- 免责声明:这个问题是基于我之前提过的另一个问题。在我最初的问题中,我手动创建了测试对象,但我不能以这种方式从我的数据库中提取它们。
$linked
和$notLinked
是节点对象的集合。根据您的应用程序,您希望它们(节点对象)执行有用的操作(使用构造函数中提供的数据)。您还可以使用类似于getData()
的方法仅返回数据。由您决定。重点是,$linked
应该包含可以从根节点访问的节点,而$notLinked
包含图中其余的节点。 - Weltschmerz