sql server如何檢索從樹中的任何孩子到祖先的所有節點？

Question

在this answer我問如何獲得所有的子節點，並且還有一個對根節點的引用。現在我意識到，我也需要相反的：

我想擁有所有的節點和所有的父母。

所以在這個簡單的樹：

1 - 2 - 3 

    L - 4 - 5 

     L - 6 

7 - 8 

我想有

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; 

（順序並不重要）

這是查詢，以獲取相反（從父母獲得所有孩子）。我試圖用它玩，但找不到解決方案。你能建議嗎？

-- 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 

Answer 1

該查詢建立鄰接表的transitive closure：所有祖先子孫對的列表。

因爲它會爲每個祖先返回所有的後代，相反的情況也是如此：對於每個後代它將返回所有的祖先。

所以這個查詢確實會返回所有可能的組合，不管遍歷順序如何：無論您是父母連接到孩子還是其他方式都無關緊要。

測試一下：

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 

Answer 2

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