如何在SQL SERVER 2005中获取给孩子的父级

Question

我有一张这样的桌子

childid      parentid
------------------------
1       0
2       1
3       2
4       2
5       3
6       4
7       0
8       7
9       8
10      1

如果我将childid设为5,那么parentid将为1(输出)

如果我将childid设为9,那么parentid将为7.(输出)

即根parentid为0,查询应该停在那里.

如何解决这样的查询？

请帮忙.

Answer 1

我认为您应该将child_id重命名为node,将parent_id重命名为child_of.您的列命名有点令人困惑

create table stack_overflow
(
node int, child_of int
);


insert into stack_overflow(node, child_of) values
(1,0),
(2,1),
(3,2),
(4,2),
(5,3),
(6,4),
(7,0),
(8,7),
(9,8),
(10,1);

这适用于任何支持CTE的RDBMS:

with find_parent(parent, child_of, recentness) as
(
    select node, child_of, 0 
    from stack_overflow
    where node = 9
    union all
    select i.node, i.child_of, fp.recentness + 1
    from stack_overflow i
    join find_parent fp on i.node = fp.child_of
)
select top 1 parent from find_parent 
order by recentness desc

输出:

parent
7

[编辑:更灵活,面向未来]:

with find_parent(node_group, parent, child_of, recentness) as
(
    select node, node, child_of, 0
    from stack_overflow
    where node in (5,9)
    union all
    select fp.node_group, i.node, i.child_of, fp.recentness + 1
    from stack_overflow i
    join find_parent fp on i.node = fp.child_of
)
select q.node_group as to_find, parent as found 
from find_parent q 
join
(
    select node_group, max(recentness) as answer
    from find_parent
    group by node_group 
) as ans on q.node_group = ans.node_group and q.recentness = ans.answer 
order by to_find    

输出:

to_find     found
5           1
9           7

如果您使用的是Postgres,则上述代码可以缩短为:

with recursive find_parent(node_group, parent, child_of, recentness) as
(
    select node, node, child_of, 0
    from stack_overflow
    where node in (5,9)
    union all
    select fp.node_group, i.node, i.child_of, fp.recentness + 1
    from stack_overflow i
    join find_parent fp on i.node = fp.child_of
)
select distinct on (node_group) node_group as to_find, parent as found 
from find_parent 
order by to_find, recentness desc

岩石上的DISTINCT!:-)