我有一张这样的 table
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,则 parentid 将为 1(输出)
如果我将 childid 设为 9,则 parentid 将为 7.(输出)
即根 parentid 是 0 并且查询应该在那里停止。
如何解决这样的查询?
请帮忙。
最佳答案
我认为您应该将 child_id 重命名为 node,将 parent_id 重命名为 child_of。你的列命名有点困惑
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:更灵活和面向 future ] :
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
如果你使用 Postgres ,上面的代码可以缩短为:
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
在岩石上与众不同! :-)
关于sql-server-2005 - 如何在 SQL SERVER 2005 中获得给定 child 的 parent ,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/1104977/