所以我有两个结构如下的表:
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。我想要它们的根节点的列表。在本例中,它将是 1、4 和 7。我如何编写查询来获取该信息?
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 的递归部分中使用 LEFT join。如果允许我执行 LEFT JOIN,下面的查询将会起作用。
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
有没有办法可以重构查询以获得我想要的结果?我无法重组数据,而且我真的更愿意远离临时表或不得不做任何昂贵的事情。
谢谢, 劳尔
最佳答案
将 LEFT JOIN 从 CTE 中移出怎么样?
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。
关于sql - T-SQL 获取层次结构中的根节点,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10045683/