24

假设我有下表：

CustomerID ParentID Name
========== ======== ====
1          null     John
2          1        James
3          2        Jenna
4          3        Jennifer
5          3        Peter
6          5        Alice
7          5        Steve
8          1        Larry

我想在一个查询中检索詹姆斯的所有后代（珍娜、詹妮弗、彼得、爱丽丝、史蒂夫）。谢谢，巴勃罗。

4

4 回答 4

34

在 SQL Server 2005 上，您可以使用CTE（公用表表达式）：

with Hierachy(CustomerID, ParentID, Name, Level)
as
(
select CustomerID, ParentID, Name, 0 as Level
    from Customers c
    where c.CustomerID = 2 -- insert parameter here
    union all
    select c.CustomerID, c.ParentID, c.Name, ch.Level + 1
    from Customers c
    inner join Hierachy ch
    on c.ParentId = ch.CustomerID
)
select CustomerID, ParentID, Name
from Hierachy
where Level > 0
于 2008-10-27T08:01:03.240 回答
3

对于自下而上，使用 mathieu 的答案并稍作修改：



with Hierachy(CustomerID, ParentID, Name, Level)
as
(
select CustomerID, ParentID, Name, 0 as Level
    from Customers c
    where c.CustomerID = 2 -- insert parameter here
    union all
    select c.CustomerID, c.ParentID, c.Name, ch.Level + 1
    from Customers c
    inner join Hierachy ch

    -- EDITED HERE --
    on ch.ParentId = c.CustomerID
    ----------------- 

)
select CustomerID, ParentID, Name
from Hierachy
where Level > 0
于 2010-09-25T13:28:29.457 回答
0

如果没有存储过程，就不能在 SQL 中进行递归。解决这个问题的方法是使用嵌套集，它们基本上将 SQL 中的树建模为一个集合。

请注意，这将需要对当前数据模型进行更改，或者可能需要弄清楚如何在原始模型上创建视图。

Postgresql 示例（使用很少的 postgresql 扩展，仅使用 SERIAL 和 ON COMMIT DROP，大多数 RDBMS 将具有类似的功能）：

设置：

CREATE TABLE objects(
    id SERIAL PRIMARY KEY,
    name TEXT,
    lft INT,
    rgt INT
);

INSERT INTO objects(name, lft, rgt) VALUES('The root of the tree', 1, 2);

添加一个孩子：

START TRANSACTION;

-- postgresql doesn't support variables so we create a temporary table that 
-- gets deleted after the transaction has finished.

CREATE TEMP TABLE left_tmp(
    lft INT
) ON COMMIT DROP; -- not standard sql

-- store the left of the parent for later use
INSERT INTO left_tmp (lft) VALUES((SELECT lft FROM objects WHERE name = 'The parent of the newly inserted node'));

-- move all the children already in the set to the right
-- to make room for the new child
UPDATE objects SET rgt = rgt + 2 WHERE rgt > (SELECT lft FROM left_tmp LIMIT 1);
UPDATE objects SET lft = lft + 2 WHERE lft > (SELECT lft FROM left_tmp LIMIT 1);

-- insert the new child
INSERT INTO objects(name, lft, rgt) VALUES(
    'The name of the newly inserted node', 
    (SELECT lft + 1 FROM left_tmp LIMIT 1), 
    (SELECT lft + 2 FROM left_tmp LIMIT 1)
);

COMMIT;

显示从下到上的轨迹：

SELECT
    parent.id, parent.lft
FROM
    objects AS current_node
INNER JOIN
    objects AS parent
ON
    current_node.lft BETWEEN parent.lft AND parent.rgt
WHERE
    current_node.name = 'The name of the deepest child'
ORDER BY
    parent.lft;

显示整个树：

SELECT
    REPEAT('   ', CAST((COUNT(parent.id) - 1) AS INT)) || '- ' || current_node.name AS indented_name
FROM
    objects current_node
INNER JOIN
    objects parent
ON
    current_node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY
    current_node.name,
    current_node.lft
ORDER BY
    current_node.lft;

从树的某个元素向下选择所有内容：

SELECT
    current_node.name AS node_name
FROM
    objects current_node
INNER JOIN
    objects parent
ON
    current_node.lft BETWEEN parent.lft AND parent.rgt
AND
    parent.name = 'child'
GROUP BY
    current_node.name,
    current_node.lft
ORDER BY
    current_node.lft;
于 2008-10-27T07:44:23.320 回答
-9

除非我遗漏了什么，否则递归是不必要的......

SELECT d.NAME FROM Customers As d
INNER JOIN Customers As p ON p.CustomerID = d.ParentID
WHERE p.Name = 'James'
于 2008-10-27T18:13:27.030 回答