使用递归返回具有唯一 ID 的重复零件

Question

我在为 SQL 数据库中的部件创建树结构时遇到一个特殊问题。 这是输出的示例，我使用缩进向父母显示孩子：

   uid: 101
   part_id: 1
   parent_part_id: NULL

       - uid: 102
       - part_id: 2
       - parent_part_id: 1

          - uid: 104
          - part_id: 4
          - parent_part_id: 2

             - uid: 106
             - part_id: 5
             - parent_part_id: 4

       - uid: 103
       - part_id: 3
       - parent_part_id: 1

          - uid: 105
          - part_id: 4
          - parent_part_id: 3

             - uid: 106
             - part_id: 5
             - parent_part_id: 4

如果parent_part_id为NULL，它是树的根。

part_id 4 的两个分支在两次出现时具有唯一的 id (uid)，因为它们链接到两个不同的父级，而part_id 4、part_id 5 的子级具有相同的 uid，因为它们链接到相同的父级。

由于我现有的系统，我需要将此数据转换为如下所示：

   uid: 101
   part_id: 1
   parent_part_id: NULL

       - uid: 102
       - part_id: 2
       - parent_part_id: 1

          - uid: 104
          - part_id: 4
          - parent_part_id: 2

             - uid: 106
             - part_id: 5
             - parent_part_id: 4

       - uid: 103
       - part_id: 3
       - parent_part_id: 1

          - uid: 105
          - part_id: 4_2
          - parent_part_id: 3

             - uid: 106
             - part_id: 5_2
             - parent_part_id: 4_2

我需要多次出现的part_id，以便在它们后面有一个数字，以便为它们提供进一步的唯一标识符，并需要它们的后续子项更改其parent_part_id以匹配它。这种情况可能会多次出现，例如“part_id 5、part_id 5_2、part_id 5_3 等”，并且子级的数量超过一层深。

如果这可以通过 MySQL 实现，我很乐意通过那里解决它，除非有更好的解决方案来通过 php 操作数据？我已经尝试通过 php 没有成功。

这是我返回数据的 MySQL：

WITH RECURSIVE generation AS (
SELECT part_id,
uid,
parent_part_id,
part_linked_order
FROM test.part_linked
WHERE
uid = 72
UNION ALL
SELECT child.part_id,
child.uid,
child.parent_part_id,
child.part_linked_order
FROM test.part_linked child
JOIN generation g
ON g.part_id = child.parent_part_id
)
SELECT
generation.part_id,
generation.uid,
generation.parent_part_id,
generation.part_linked_order
FROM generation
Order By
part_linked_order ASC;

（我使用part_linked_order来确定在其分支中显示哪些订单零件，但我不认为它与问题相关）

我尝试使用递归循环遍历重复部分的子项以正确连接它们，但到目前为止还没有效果。

这是输出的数组，以便于测试：

$array = array(
  array(
    "uid" => 101,
    "part_id" => 1,
    "parent_part_id" => NULL
  ),
  array(
    "uid" => 102,
    "part_id" => 2,
    "parent_part_id" => 1
  ),
  array(
    "uid" => 104,
    "part_id" => 4,
    "parent_part_id" => 2
  ),
  array(
    "uid" => 106,
    "part_id" => 5,
    "parent_part_id" => 4
  ),
  array(
    "uid" => 103,
    "part_id" => 3,
    "parent_part_id" => 1
  ),
  array(
    "uid" => 105,
    "part_id" => 4,
    "parent_part_id" => 3
  ),
  array(
    "uid" => 106,
    "part_id" => 5,
    "parent_part_id" => 4
  )
);

我想要输出的内容：

$array = array(
  array(
    "uid" => 101,
    "part_id" => 1,
    "parent_part_id" => NULL
  ),
  array(
    "uid" => 102,
    "part_id" => 2,
    "parent_part_id" => 1
  ),
  array(
    "uid" => 104,
    "part_id" => 4,
    "parent_part_id" => 2
  ),
  array(
    "uid" => 106,
    "part_id" => 5,
    "parent_part_id" => 4
  ),
  array(
    "uid" => 103,
    "part_id" => 3,
    "parent_part_id" => 1
  ),
  array(
    "uid" => 105,
    "part_id" => 4_2,
    "parent_part_id" => 3
  ),
  array(
    "uid" => 106,
    "part_id" => 5_2,
    "parent_part_id" => 4_2
  )
);

表结构

• 主键：uid
• part_id（字符串），
• parent_part_id（字符串），
• part_linked_order（int）

Answer 1

我相信您所需要的只是添加行号计算，然后有条件地连接part_id和该行号以获得想要的结果：

WITH RECURSIVE generation AS (
    SELECT part_id
        , uid
        , parent_part_id
    FROM part_linked
    WHERE uid = 72
    
    UNION ALL
    
    SELECT child.part_id
        , child.uid
        , child.parent_part_id
    FROM part_linked child
    JOIN generation g ON g.part_id = child.parent_part_id
    )
SELECT
      case when p_order > 1 then concat(part_id,'_',p_order) else part_id end as part_linked_order
    , part_id
    , uid
    , parent_part_id
FROM (
    SELECT
          generation.part_id
        , generation.uid
        , generation.parent_part_id
        , ROW_NUMBER() OVER (PARTITION BY part_id ORDER BY parent_part_id) p_order
    FROM generation
  ) d
ORDER BY part_linked_order ASC;

部分链接订单	part_id	流体	parent_part_id
1	1	72	空
2	2	102	1
3	3	103	1
4	4	104	2
4_2	4	105	3
5	5	106	4
5_2	5	106	4
5_3	5	106	4
5_4	5	106	4

看到这个MYsql fiddle