将邻接列表层次结构展平为所有路径的列表

Question

我有一个表使用Adjacency List模型存储分层信息.(使用自引用键 - 下面的示例.此表可能看起来很熟悉):

category_id name                 parent
----------- -------------------- -----------
1           ELECTRONICS          NULL
2           TELEVISIONS          1
3           TUBE                 2
4           LCD                  2
5           PLASMA               2
6           PORTABLE ELECTRONICS 1
7           MP3 PLAYERS          6
8           FLASH                7
9           CD PLAYERS           6
10          2 WAY RADIOS         6

---

将上述数据"压扁"成这样的东西的最佳方法是什么？

category_id lvl1        lvl2        lvl3        lvl4
----------- ----------- ----------- ----------- -----------
1           1           NULL        NULL        NULL
2           1           2           NULL        NULL
6           1           6           NULL        NULL
3           1           2           3           NULL
4           1           2           4           NULL
5           1           2           5           NULL
7           1           6           7           NULL
9           1           6           9           NULL
10          1           6           10          NULL
8           1           6           7           8

---

每行是通过层次结构的一个"路径",除了每个节点(不仅是每个叶节点)都有一行.category_id列表示当前节点,"lvl"列表示其祖先.当前节点的值也必须位于最右边的lvl列中.lvl1列中的值将始终表示根节点,lvl2中的值将始终表示lvl1的直接后代,依此类推.

如果可能,生成此输出的方法将在SQL中,并且适用于n层层次结构.

Answer 1

在一个简单的邻接列表中进行多级查询总是涉及自左连接.制作一个右对齐的表格很容易:

SELECT category.category_id,
    ancestor4.category_id AS lvl4,
    ancestor3.category_id AS lvl3,
    ancestor2.category_id AS lvl2,
    ancestor1.category_id AS lvl1
FROM categories AS category
    LEFT JOIN categories AS ancestor1 ON ancestor1.category_id=category.category_id
    LEFT JOIN categories AS ancestor2 ON ancestor2.category_id=ancestor1.parent
    LEFT JOIN categories AS ancestor3 ON ancestor3.category_id=ancestor2.parent
    LEFT JOIN categories AS ancestor4 ON ancestor4.category_id=ancestor3.parent;

像你的例子一样左对齐它有点棘手.想到这一点:

SELECT category.category_id,
    ancestor1.category_id AS lvl1,
    ancestor2.category_id AS lvl2,
    ancestor3.category_id AS lvl3,
    ancestor4.category_id AS lvl4
FROM categories AS category
    LEFT JOIN categories AS ancestor1 ON ancestor1.parent IS NULL
    LEFT JOIN categories AS ancestor2 ON ancestor1.category_id<>category.category_id AND ancestor2.parent=ancestor1.category_id
    LEFT JOIN categories AS ancestor3 ON ancestor2.category_id<>category.category_id AND ancestor3.parent=ancestor2.category_id
    LEFT JOIN categories AS ancestor4 ON ancestor3.category_id<>category.category_id AND ancestor4.parent=ancestor3.category_id
WHERE
    ancestor1.category_id=category.category_id OR
    ancestor2.category_id=category.category_id OR
    ancestor3.category_id=category.category_id OR
    ancestor4.category_id=category.category_id;

适用于n层层次结构.

抱歉,在邻接列表模型中无法进行任意深度查询.如果您经常进行此类查询,则应将模式更改为存储分层信息的其他模型之一:完全邻接关系(存储所有祖先 - 后代关系),物化路径或嵌套集.

如果类别不会移动很多(这通常就像你的例子那样的商店),我倾向于嵌套集.

Answer 2

如前所述,SQL没有干净的方法来实现具有动态变化的列数的表.我之前使用的唯一两个解决方案是:1.固定数量的自连接,给出固定数量的列(每个BobInce的AS)2.在单个列中将结果生成为字符串

第二个听起来很怪诞; 将ID存储为字符串？!但是当输出格式化为XML或其他东西时,人们似乎并不介意这么多.

同样,如果您想要在SQL中加入结果,这几乎没用.如果要将结果提供给应用程序,则它可能非常合适.但就个人而言,我更喜欢在应用程序而不是SQL中进行展平

我被困在一个10英寸的屏幕上,无法访问SQL,因此我无法提供经过测试的代码,但基本方法是以某种方式利用递归;
- 递归标量函数可以做到这一点
- MS SQL可以使用递归WITH语句执行此操作(更高效)

标量函数(类似):

CREATE FUNCTION getGraphWalk(@child_id INT)
RETURNS VARCHAR(4000)
AS
BEGIN

  DECLARE @graph VARCHAR(4000)

  -- This step assumes each child only has one parent
  SELECT
    @graph = dbo.getGraphWalk(parent_id)
  FROM
    mapping_table
  WHERE
    category_id = @child_id
    AND parent_id IS NOT NULL

  IF (@graph  IS NULL)
    SET @graph = CAST(@child_id AS VARCHAR(16))
  ELSE
    SET @graph = @graph + ',' + CAST(@child_id AS VARCHAR(16))

  RETURN @graph

END


SELECT
  category_id                         AS [category_id],
  dbo.getGraphWalk(category_id)       AS [graph_path]
FROM
  mapping_table
ORDER BY
  category_id

我有一段时间没有使用递归的WITH,但即使我没有SQL来测试任何东西,我也会给出语法.

递归WITH

WITH
  result (
    category_id,
    graph_path
  )
AS
(
  SELECT
    category_id,
    CAST(category_id AS VARCHAR(4000))
  FROM
    mapping_table
  WHERE
    parent_id IS NULL

  UNION ALL

  SELECT
    mapping_table.category_id,
    CAST(result.graph_path + ',' + CAST(mapping_table.category_id AS VARCHAR(16)) AS VARCHAR(4000))
  FROM
    result
  INNER JOIN
    mapping_table
      ON result.category_id = mapping_table.parent_id
)

SELECT
  *
FROM
  result
ORDER BY
  category_id

编辑 - 两者的输出是相同的:

1   '1'
2   '1,2'
3   '1,2,3'
4   '1,2,4'
5   '1,2,5'
6   '1,6'
7   '1,6,7'
8   '1,6,7,8'
9   '1,6,9'