SQL开发人员,我有一个计划不周的数据库作为了解SQL Server 2012的重要任务。
所以,有表Elem
:
+-----------+----+---+----------+------------+
|VERSION(PK)|NAME|KEY|PARENT_KEY|DIST_KEY(FK)|
+-----------+----+---+----------+------------+
|1 |a |12 |NULL |1 |
+-----------+----+---+----------+------------+
|2 |b |13 |12 |1 |
+-----------+----+---+----------+------------+
|3 |c |14 |13 |1 |
+-----------+----+---+----------+------------+
|4 |d |15 |12 |1 |
+-----------+----+---+----------+------------+
|5 |e |16 |NULL |1 |
+-----------+----+---+----------+------------+
|6 |e |17 |NULL |2 |
+-----------+----+---+----------+------------+
更新行后,我需要检查元素的父键,不允许元素是自己的老奶奶......
当我删除该行时,我需要删除所有儿童和孩子的孩子等等。
问题是:
如何选择DIST的一个元素的所有“父+祖父母+等”?
如何选择DIST的一个元素的所有“儿子+孙子+等”?
我读过有关CTE的解决方案,但我没有元素根,我甚至无法理解如何使用CTE。
请帮忙!
感谢。
答案 0 :(得分:24)
我遇到了这个问题,我通过这种方式解决了问题
--all "parent + grandparent + etc" @childID Replaced with the ID you need
with tbParent as
(
select * from Elem where [KEY]=@childID
union all
select Elem.* from Elem join tbParent on Elem.[KEY]=tbParent.PARENT_KEY
)
SELECT * FROM tbParent
--all "sons + grandsons + etc" @parentID Replaced with the ID you need
with tbsons as
(
select * from Elem where [KEY]=@parentID
union all
select Elem.* from Elem join tbsons on Elem.PARENT_KEY=tbsons.[KEY]
)
SELECT * FROM tbsons
PS.My英语不好。
答案 1 :(得分:1)
这是一个递归查询,为您提供元素的所有祖先和所有后代。根据情况一起使用或分开使用。替换where子句以获得所需的记录。在这个例子中,我正在寻找键13(这是名称= b的元素)并找到它的祖先12 / a及其后代14 / c。
with all_ancestors(relation, version, name, elem_key, parent_key, dist_key)
as
(
-- the record itself
select 'self ' as relation, self.version, self.name, self.elem_key, self.parent_key, self.dist_key
from elem self
where elem_key = 13
union all
-- all its ancestors found recursively
select 'ancestor ' as relation, parent.version, parent.name, parent.elem_key, parent.parent_key, parent.dist_key
from elem parent
join all_ancestors child on parent.elem_key = child.parent_key
)
, all_descendants(relation, version, name, elem_key, parent_key, dist_key)
as
(
-- the record itself
select 'self ' as relation, self.version, self.name, self.elem_key, self.parent_key, self.dist_key
from elem self
where elem_key = 13
union all
-- all its descendants found recursively
select 'descendant' as relation, child.version, child.name, child.elem_key, child.parent_key, child.dist_key
from elem child
join all_descendants parent on parent.elem_key = child.parent_key
)
select * from all_ancestors
union
select * from all_descendants
order by elem_key
;
这是SQL小提琴:http://sqlfiddle.com/#!6/617ee/28。
答案 2 :(得分:0)
我已经创建了一个函数来查找特定孩子的父母,你必须传递孩子的Id。
这将以逗号分隔的字符串形式返回父列表。 试试这个,如果它适合你。
我假设parent_key with null value
是root。
CREATE FUNCTION checkParent(@childId INT)
RETURNS VARCHAR(MAX)
AS
BEGIN
DECLARE @parentId VARCHAR(MAX) = NULL
DECLARE @parentKey INT = null
SET @parentId = (SELECT parent_key FROM Elem WHERE [KEY] = @childId)
WHILE(@parentKey IS NOT NULL)
begin
SET @parentId = @parentId + ', ' + (SELECT parent_key FROM Elem WHERE [KEY] = @parentId)
SET @parentKey = (SELECT parent_key FROM Elem WHERE [KEY] = @parentId)
END
RETURN @parentId
END
GO
答案 3 :(得分:0)
我不认为它可以在一个选择中完成,所以你可以选择所有的父母,祖父母,....一种方法是将elem表连接到自己,这取决于你做多少级别的连接,你会得到多少个孩子,孙子。
解决方案可以像这样思考(对于第二种情况)
这将选择所有父母,孩子和孙子
Select
parent.key as parent_key,
child.key as child_key,
grandchild.key as grandchild_key
from elem parent
join elem child on (elem.key=child.parentkey)
join elem grandchild on (child.key=grandchild.parentkey)
where parent.parentkey is null; -- this make you sure that first level will be parents
第一种情况的解决方案只是你将连接不是'key = parentkey'风格但是oposite'parentkey = key'的表。