[英]How do I create a recursive query in MSSQL 2005?
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James的所有后代(Jenna,Jennifer,Peter,Alice,Steve)。 谢谢,巴勃罗。
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
自下而上使用mathieu的答案稍作修改:
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
没有存储过程,您无法在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中的树建模为一组。
请注意,这将需要更改当前数据模型或可能确定如何在原始模型上创建视图。
Postgresql示例(使用很少的postgresql扩展,只有SERIAL和ON COMMIT DROP,大多数RDBMS将具有类似的功能):
设定:
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;
从下到上显示路径:
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;
从树的某个元素中选择所有内容:
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;
除非我遗漏了什么,否则递归不是必要的......
SELECT d.NAME FROM Customers As d
INNER JOIN Customers As p ON p.CustomerID = d.ParentID
WHERE p.Name = 'James'
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