问题描述
我有一个树形层次结构,它看起来是内置在表中的,并且parent_id指向前一个根节点.
I have a tree hierarchy look this built into a table with the parent_id pointing to the previous root node.
我正在遍历所有根节点(root1,root2),并且将root1和child1的路径设置为root1或root1/child1.为了找到child1的路径,我将至少要调用2才能形成路径.有没有一种有效的方法来填充路径,因为我们处理了嵌套在5-7层深处的大量根节点和子节点.
I am iterating through all root nodes (root1, root2) and I am setting path to either root1 or root1/child1 for root1 and child1. In order to find the path for child1, I will have to make at-least 2 calls to form the path. Is there an efficient way to fill the path, since we deal with a very large number of root nodes and children which are nested 5-7 levels deep.
create table foo (id, name, parent_id, path)
insert into foo (1, "root1', null, null)
insert into foo (2, "child1', 1, null)
root1 (path = null)
child1 (path = root1)
subchild1 (path = root1/child1)
root2
child2
subchild2
推荐答案
您可以使用问题中提到的存储过程,因为嵌套最多可以达到7层.
You can go with a stored procedure as you have mentioned in your question as the nesting can be up to 7 level deep.
CREATE PROCEDURE updatePath()
BEGIN
declare cnt, n int;
select count(*) into n from foo where parent_id is null;
update foo a, foo b set a.path = b.name where b.parent_id is null and a.parent_id = b.id;
select count(*) into cnt from foo where path is null;
while cnt > n do
update foo a, foo b set a.path = concat(b.path, '/', b.name) where b.path is not null and a.parent_id = b.id;
select count(*) into cnt from foo where path is null;
end while;
END//
要检查实际记录,我们只在路径列中打印了具有空值的纯记录
To check the actual record we just printed the plain records having null value in path column
select * from foo
结果:
| ID | NAME | PARENT_ID | PATH |
------------------------------------------
| 1 | root1 | (null) | (null) |
| 2 | child1 | 1 | (null) |
| 3 | subchild1 | 2 | (null) |
| 4 | child2 | 1 | (null) |
| 5 | child3 | 1 | (null) |
| 6 | subchild2 | 4 | (null) |
| 7 | subsubchild1 | 6 | (null) |
调用过程:
call updatepath
过程执行后的结果:
select * from foo
结果:
| ID | NAME | PARENT_ID | PATH |
----------------------------------------------------------
| 1 | root1 | (null) | (null) |
| 2 | child1 | 1 | root1 |
| 3 | subchild1 | 2 | root1/child1 |
| 4 | child2 | 1 | root1 |
| 5 | child3 | 1 | root1 |
| 6 | subchild2 | 4 | root1/child2 |
| 7 | subsubchild1 | 6 | root1/child2/subchild2 |
SQLFIDDLE
希望这会有所帮助....
SQLFIDDLE
Hope this helps....
这篇关于如何找到树表示的层次结构路径的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!