3

在这个答案中，我询问了如何获取所有子节点以及对根节点的引用。现在我意识到我也需要相反的：

我想要所有的节点和所有的父母。

所以在这个简单的树中：

1 - 2 - 3

    L - 4 - 5

        L - 6

7 - 8

我想拥有

1 1;
2 2;
2 1;
3 3;
3 2;
3 1;
4 4;
4 2;
4 1;
5 5;
5 4;
5 2;
5 1;
6 6;
6 4;
6 2;
6 1;
7 7;
8 8;
8 7;

（顺序不重要）

这是获取相反结果的查询（从父级获取所有子级）。我试图玩它，但找不到解决方案。你能建议吗？

-- get all childs of all parents
WITH    q AS
        (
        SELECT  ID_CUSTOMER, ID_CUSTOMER AS root_customer
        FROM    CUSTOMERS c
        UNION ALL
        SELECT  c.ID_CUSTOMER, q.root_customer
        FROM    q
        JOIN    CUSTOMERS c 
        ON      c.ID_PARENT_CUSTOMER = q.ID_CUSTOMER
        )
SELECT  *
FROM    q
4

2 回答 2

3

此查询构建transitive closure邻接列表：所有祖先-后代对的列表。

由于它将返回每个祖先的所有后代，反之亦然：对于每个后代，它将返回其所有祖先。

因此，无论遍历顺序如何，这个查询都会返回所有可能的组合：无论您是连接父母还是孩子，都无关紧要。

让我们测试一下：

WITH    customers (id_customer, id_parent_customer) AS
        (
        SELECT  *
        FROM    (
                VALUES  (1, NULL),
                        (2, 1),
                        (3, 2),
                        (4, 2),
                        (5, 4),
                        (6, 4),
                        (7, NULL),
                        (8, 7)
                ) t (a, b)
        ),
        q AS
        (
        SELECT  ID_CUSTOMER, ID_CUSTOMER AS root_customer
        FROM    CUSTOMERS c
        UNION ALL
        SELECT  c.ID_CUSTOMER, q.root_customer
        FROM    q
        JOIN    CUSTOMERS c 
        ON      c.ID_PARENT_CUSTOMER = q.ID_CUSTOMER
        )
SELECT  *
FROM    q
ORDER BY
        id_customer, root_customer DESC
于 2011-04-19T13:34:32.917 回答
0 
with q (
select id_customer, id_parent_customer from customers
union all
select id_customer, id_parent_customer from customers
join q on customers.id_parent_customer = q.id_customer
) select * from q
于 2011-04-19T13:24:44.377 回答