修改后的先序树遍历:选择1层深度的节点

发布于 2024-12-08 13:32:38 字数 839 浏览 7 评论 0原文

我使用修改后的先序树遍历算法保存了分层有序数据。

这是表格内容:

id  lft  rgt  name
1   1    10   topnode
2   2    3    level1
3   4    7    level1
4   5    6    level2
5   8    9    level1

可视化:

Scheme

我想要的是仅选择某个节点的子节点(所以不是子节点)。让我们说“顶部节点”。我正在尝试修复一个查询,但我似乎无法理解它。

搜索互联网给我带来了一些时间,例如:我可以计算每个节点的深度,但我似乎无法对其进行选择。

此查询

SELECT node.*, (COUNT(parent.id) - 1) AS depth
FROM tree AS node
CROSS JOIN tree AS parent
WHERE (node.lft BETWEEN parent.lft AND parent.rgt)
GROUP BY node.id
ORDER BY node.lft

显示每个节点的深度:

id  lft  rgt  name     depth
1   1    10   topnode  0
2   2    3    level1   1
3   4    7    level1   1
4   5    6    level2   2
5   8    9    level1   1

这很好,但我不能使用列深度作为条件!

I have hierarchical ordered data saved using the modified preorder tree traversal algorithm.

Here's tables content:

id  lft  rgt  name
1   1    10   topnode
2   2    3    level1
3   4    7    level1
4   5    6    level2
5   8    9    level1

Visualised:

Scheme

What I want is to select just the childnodes of a certain node (so not the childnodes of the childnodes). Let's say 'topnode'. I'm trying to fix a query, but I can't seem to get my head around it.

Searching the internet brings me a while, for example: I can calculate the depth of each node, but I just can't seem to select on it.

This query

SELECT node.*, (COUNT(parent.id) - 1) AS depth
FROM tree AS node
CROSS JOIN tree AS parent
WHERE (node.lft BETWEEN parent.lft AND parent.rgt)
GROUP BY node.id
ORDER BY node.lft

shows the depth of each node:

id  lft  rgt  name     depth
1   1    10   topnode  0
2   2    3    level1   1
3   4    7    level1   1
4   5    6    level2   2
5   8    9    level1   1

That's great, but I can't use the column depth as a condition!

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评论(1

江南月 2024-12-15 13:32:38

我认为这应该可以解决问题

SELECT node.*, (COUNT(parent.id) - 1) AS depth
FROM tree AS node
CROSS JOIN tree AS parent
WHERE (node.lft BETWEEN parent.lft AND parent.rgt)
GROUP BY node.id
HAVING depth = {{ENTER YOUR REQUIRED DEPTH HERE}}
ORDER BY node.lft

希望有帮助

I think this should do the trick

SELECT node.*, (COUNT(parent.id) - 1) AS depth
FROM tree AS node
CROSS JOIN tree AS parent
WHERE (node.lft BETWEEN parent.lft AND parent.rgt)
GROUP BY node.id
HAVING depth = {{ENTER YOUR REQUIRED DEPTH HERE}}
ORDER BY node.lft

Hope that helps

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