SQL

Question

我有以下 table "x" 层级属性：

+-----------+--------+--------------+-----------+-------------+
| row       | value  | displayvalue | parentrow | parentvalue |
+-----------+--------+--------------+-----------+-------------+
| HIE1      | VALUE3 | value 3      | NULL      | NULL        |
| HIE2      | VALUE1 | value 11     | HIE1      | VALUE1      |
| HIE2      | VALUE1 | value 21     | HIE1      | VALUE2      |
| HIE2      | VALUE1 | value 31     | HIE1      | VALUE3      |
| HIE3      | VALUE1 | value 111    | HIE2      | VALUE1      |
| HIE3      | VALUE1 | value 311    | HIE2      | VALUE1      |
| HIE3      | VALUE1 | value 221    | HIE2      | VALUE2      |
+-----------+--------+--------------+-----------+-------------+

这只是一个示例 - 层次结构可能更大或更小，并且行代码不必是数字。我该如何提取这个 table 中唯一正确的分支？即：

+-----------+--------+--------------+-----------+-------------+
| row       | value  | displayvalue | parentrow | parentvalue |
+-----------+--------+--------------+-----------+-------------+
| HIE1      | VALUE3 | value 3      | NULL      | NULL        |
| HIE2      | VALUE1 | value 31     | HIE1      | VALUE3      |
| HIE3      | VALUE1 | value 311    | HIE2      | VALUE1      |
+-----------+--------+--------------+-----------+-------------+

这需要递归，我试过了：

With Attrs AS (
    SELECT "row",
           "value",
           "displayvalue",
           "parentrow",
           "parentvalue",
     FROM x

    UNION ALL

    SELECT child."row",
           child."value",
           child."displayvalue",
           child."parentrow",
           child."parentvalue",
     FROM x child
     JOIN Attrs ON child."parentrow" = Attrs."row" AND child."parentvalue" = Attrs."value"
     WHERE Attrs."parentrow" IS NOT NULL

  )
  SELECT "row", "value", "displayvalue", "parentrow", "parentvalue"
  FROM Attrs;

但这似乎扩大了结果池，而不是缩小了范围。我做错了什么？

编辑

忘记在 table 中添加序列列。这些是额外的条目：

+-----------+--------+-----------+
| row       | seqno  | parentseq |
+-----------+--------+-----------+
| HIE1      | 3      | NULL      |
| HIE2      | 10     | 1         |
| HIE2      | 13     | 2         |
| HIE2      | 11     | 3         |
| HIE3      | 15     | 10        |
| HIE3      | 16     | 11        |
| HIE3      | 17     | 12        |
+-----------+--------+-----------+

所需的行将具有：

+-----------+--------+-----------+
| row       | seqno  | parentseq |
+-----------+--------+-----------+
| HIE1      | 3      | NULL      |
| HIE2      | 11     | 3         |
| HIE3      | 16     | 11        |
+-----------+--------+-----------+

Answer 1

您可以使用以下查询：

;WITH Attrs AS (
   -- Anchor query: Get the root records
   SELECT row, value, displayvalue, parentrow, parentvalue
   FROM x
   WHERE parentrow IS NULL

   UNION ALL

   -- Recursive query: Get the records of the next level
   SELECT child.row, child.value, child.displayvalue, child.parentrow, child.parentvalue
   FROM x AS child
   INNER JOIN Attrs AS parent 
      ON child.parentrow = parent.row AND child.parentvalue = parent.value
)
SELECT row, value, displayvalue, parentrow, parentvalue
FROM Attrs

递归 CTE 的第一个子查询，也称为 anchor 查询，获取层次结构的根记录。因此，您必须在此子查询上放置 parentrow IS NULL。

第二个子查询将 table 与上一个递归的结果连接起来，以获取树层次结构中下一级的记录。不需要在此子查询中使用 parentrow IS NULL，因为第一次执行递归产生的记录已经根节点记录。

Demo here

编辑： 将序列号添加到我们得到的查询中：

;WITH Attrs AS (
   -- Anchor query: Get the root records
   SELECT row, seqno, value, displayvalue, 
          parentrow, parentvalue, parentseq
   FROM x
   WHERE parentrow IS NULL

   UNION ALL

   -- Recursive query: Get the records of the next level
   SELECT child.row, child.seqno, child.value, child.displayvalue, 
          child.parentrow, child.parentvalue, child.parentseq
   FROM x AS child
   INNER JOIN Attrs AS parent 
      ON child.parentrow = parent.row AND 
         child.parentvalue = parent.value AND 
         child.parentseq = parent.seqno
)
SELECT row, value, displayvalue, parentrow, parentvalue
FROM Attrs

SQL - 获取层次结构的单个分支

SQL - Getting a single branch of a hierarchy

sql-server

recursion

tree-traversal

common-table-expression