SQL获得最低水平的孩子，根节点

我有数据库架构：[ID]，[ParrentId]，[更多的表]SQL获得最低水平的孩子，根节点

我有层次，如：

1. a 
2. aa 
    3. aaa_1 
    3. aaa_2 
1. b 
2. bb 
1. c 
2. cc 
    3. ccc_1 
    4. cccc 
    3. ccc_2

我想（选择*其中X）=> [X，最低级别子]例如： [a，aaa_1] [a，aaa_2]; [CC，CCCC]等

我可以

SELECT t1.name FROM 
category AS t1 LEFT JOIN category as t2 
ON t1.category_id = t2.parent 
WHERE t2.category_id IS NULL;

得到最低的孩子，但我不知道如何与根节点加入。

来源

2016-01-22 therko

哪些DBMS您使用的？ Postgres的？甲骨文？ –

我建议添加一个反映层级的字段。然后使用以下SQL：SELECT name，max（level）FROM category WHERE parent = {parent_的category_id} – ahPo

考虑：

数据库管理系统是SQL服务器;
树的最高级别节点有parent = NULL;
你想要所有层次的树木的所有最低的叶子，而不仅仅是根;
您希望将所有节点放在最底层，而不只是一个;

此查询将做到这一点：

WITH r (category_id, name, root, depth) 
-- finds the root relationship 
AS (
    SELECT category_id, name, category_id, 0 
     FROM category 
     -- WHERE parent IS NULL -- this would only look at root nodes 
    UNION ALL 
    SELECT c.category_id, c.name, r.root, r.depth + 1 
     FROM r 
     JOIN category c 
      ON c.parent = r.category_id 
), s (category_id, name, root, window_id) 
-- finds the lowest leaves 
AS (
    SELECT category_id, name, root, RANK() OVER(partition by root order by depth DESC) 
     FROM r 
) 
SELECT c.name AS NodeName, s.Name AS DeepLeafName 
    FROM category c 
    JOIN s 
     ON c.category_id = s.root 
     WHERE s.window_id = 1;

下面是结果集：

来源

2016-01-22 16:08:41

使用SQL Server，你可以试试这个：

With CTE as 
(
Select ID as Child, lev = 1 
from category 
where ID = X 
UNION ALL 
Select category.ID, CTE.lev + 1 
from category 
    inner join CTE ON category.ParentID = CTE.Child 
) 

select CTE_1.Child, CTE_2.Child 
from CTE as CTE_1 
    inner join CTE as CTE_2 
where CTE_1.lev = 1 AND CTE_2.lev = (select MAX(CTE.lev) from CTE)

来源

2016-01-22 15:59:58 Polux2

SQL获得最低水平的孩子，根节点

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