问题描述

所以我有两个这样结构的表:

So I have two tables structured like so:

CREATE TABLE #nodes(node int NOT NULL);
ALTER TABLE #nodes ADD CONSTRAINT PK_nodes PRIMARY KEY CLUSTERED (node);

CREATE TABLE #arcs(child_node int NOT NULL, parent_node int NOT NULL);
ALTER TABLE #arcs ADD CONSTRAINT PK_arcs PRIMARY KEY CLUSTERED (child_node, parent_node);

INSERT INTO #nodes(node)
VALUES (1), (2), (3), (4), (5), (6), (7);

INSERT INTO #arcs(child_node, parent_node)
VALUES (2, 3), (3, 4), (2, 6), (6, 7);

如果我有两个节点，比如说 1 和 2.我想要一个它们的根节点的列表.在这种情况下，它将是 1、4 和 7.我如何编写查询来获取该信息?

If I have two nodes, lets say 1 and 2. I want a list of their root nodes. In this case it would be 1, 4, and 7. How can I write a query to get me that information ?

我尝试编写它，但遇到了一个问题，即由于某种未知原因，我无法在 CTE 的递归部分使用 LEFT 连接.如果允许我执行 LEFT JOIN，这里是可行的查询.

I took a stab at writing it but ran into the issue that I can't use a LEFT join in the recursive part of a CTE for some unknown reason. Here is the query that would work if I was allowed to do a LEFT JOIN.

WITH root_nodes
AS (
    -- Grab all the leaf nodes I care about and their parent
    SELECT n.node as child_node, a.parent_node
    FROM #nodes n
    LEFT JOIN #arcs a
      ON n.node = a.child_node
    WHERE n.node IN (1, 2)

    UNION ALL

    -- Grab all the parent nodes
    SELECT rn.parent_node as child_node, a.parent_node
    FROM root_nodes rn
    LEFT JOIN #arcs a -- <-- LEFT JOINS are Illegal for some reason :(
      ON rn.parent_node = a.child_node
    WHERE rn.parent_node IS NOT NULL
)
SELECT DISTINCT rn.child_node as root_node
FROM root_nodes rn
WHERE rn.parent_node IS NULL

有什么方法可以重组查询以获得我想要的?我无法重组数据，我真的希望远离临时表或不得不做任何昂贵的事情.

Is there a way I can restructure the query to get what I want ? I can't restructure the data and I would really prefer to stay away from temporary tables or having to do anything expensive.

谢谢，劳尔

推荐答案

将 LEFT JOIN 移出 CTE 怎么样?

How about moving the LEFT JOIN out of the CTE?

WITH root_nodes
AS (
    -- Grab all the leaf nodes I care about
    SELECT NULL as child_node, n.node as parent_node
    FROM #nodes n
    WHERE n.node IN (1, 2)

    UNION ALL

    -- Grab all the parent nodes
    SELECT rn.parent_node as child_node, a.parent_node
    FROM root_nodes rn
        JOIN #arcs a
      ON rn.parent_node = a.child_node
)
SELECT DISTINCT rn.parent_node AS root_node
FROM root_nodes rn
    LEFT JOIN #arcs a
  ON rn.parent_node = a.child_node
WHERE a.parent_node IS NULL

结果集为 1、4、7.

The result set is 1, 4, 7.

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