如何获得PostgreSQL中最顶级的父亲

Question

我有一个树型结构表，列中有： id，parent，name。 给定一棵树A-> B-> C， 我怎么能根据C的ID得到最顶级的父A的ID？ 尤其是如何用“递归”编写SQL？ 谢谢！

Answer 1

要实现递归查询，您需要一个Common Table Expression（CTE）。 该查询计算所有父节点的祖先。既然我们只想要最高级别，我们选择level = 0的地方。

WITH RECURSIVE Ancestors AS 
(
    SELECT id, parent, 0 AS level FROM YourTable WHERE parent IS NULL 
    UNION ALL 
    SELECT child.id, child.parent, level+1 FROM YourTable child INNER JOIN 
     Ancestors p ON p.id=child.parent 

) 
SELECT * FROM Ancestors WHERE a.level=0 AND a.id=C 

如果您想要获取所有数据，请使用ID上的内连接，例如，

SELECT YourTable.* FROM Ancestors a WHERE a.level=0 AND a.id=C 
    INNER JOIN YourTable ON YourTable.id = a.id 

Answer 2

WITH RECURSIVE q AS 
     (
     SELECT m 
     FROM mytable m 
     WHERE id = 'C' 
     UNION ALL 
     SELECT m 
     FROM q 
     JOIN mytable m 
     ON  m.id = q.parent 
     ) 
SELECT (m).* 
FROM q 
WHERE (m).parent IS NULL