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问题:
I've been using the crap out of the Nested Set Model lately. I have enjoyed designing queries for just about every useful operation and view. One thing I'm stuck on is how to select the immediate children (and only the children, not further descendants!) of a node.
To be honest, I do know of a way - but it involves unmanageable amounts of SQL. I'm sure there is a more straightforward solution.
回答1:
Did you read the article you posted? It's under the heading "Find the Immediate Subordinates of a Node"
SELECT node.name, (COUNT(parent.name) - (sub_tree.depth + 1)) AS depth
FROM nested_category AS node,
nested_category AS parent,
nested_category AS sub_parent,
(
SELECT node.name, (COUNT(parent.name) - 1) AS depth
FROM nested_category AS node,
nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
AND node.name = 'PORTABLE ELECTRONICS'
GROUP BY node.name
ORDER BY node.lft
)AS sub_tree
WHERE node.lft BETWEEN parent.lft AND parent.rgt
AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
AND sub_parent.name = sub_tree.name
GROUP BY node.name
HAVING depth <= 1
ORDER BY node.lft;
However, what I do (this is cheating) is I combined the nested set with adjacency lists -- I embed a "parent_id" in the table, so I can easily ask for the children of a node.
回答2:
It seems to me this should be easily doable without the subqueries or parent column redundancy! For example, given parent's left and right are already known:
SELECT child.id
FROM nodes AS child
LEFT JOIN nodes AS ancestor ON
ancestor.left BETWEEN @parentleft+1 AND @parentright-1 AND
child.left BETWEEN ancestor.left+1 AND ancestor.right-1
WHERE
child.left BETWEEN @parentleft+1 AND @parentright-1 AND
ancestor.id IS NULL
That is, “from all descendents of the node in question, pick ones with no ancestor between themselves and the node”.
回答3:
THIS ONE IS BETTER AND SMALLER
User "bobince" almost had it. I figured it out and got it to work for me because I have a little more MySQL experience than most. However, I can see why bobince's answer might scare people off. His query is incomplete. You need to select the parent_left and parent_right into mysql variables first.
The two queries below assume that your table is named tree
, your left column is named lft
, right column is named rgt
, and that your primary key is named id
. Change these values to suit your needs. Also, examine the first select statement. You will see that I am looking up the immediate descendants of node 5. Change the number 5 to look for children of whatever node you want.
I personally think this is a sleeker, sexier, and more efficient query than the others presented so far.
SELECT `lft`, `rgt` INTO @parent_left, @parent_right FROM efm_files WHERE `id` = 5;
SELECT `child`.`id`
FROM `tree` AS `child`
LEFT JOIN `tree` AS `ancestor` ON
`ancestor`.`lft` BETWEEN @parent_left+1 AND @parent_right-1 AND
`child`.`lft` BETWEEN `ancestor`.`lft`+1 AND `ancestor`.`rgt`-1
WHERE
`child`.`lft` BETWEEN @parent_left+1 AND @parent_right-1 AND
`ancestor`.`id` IS NULL
回答4:
i know im doing a necro post,
but here's my opinion.
why not include a "depth" column in your nested set?
the depth column will indicate the "level" of an item.
so, to select the immediate childs of an item, just do
select c.*
from tree as p
join tree as c on (c.left > p.left and c.right < p.right and c.depth = p.dept + 1)
where p.id = @parentID
回答5:
I'd go with a depth column, too. But use
SELECT Child.Node, Child.LEFT, Child.RIGHT
FROM Tree AS Child, Tree AS Parent
WHERE
Child.Depth = Parent.Depth + 1
AND Child.LEFT > Parent.LEFT
AND Child.RIGHT < Parent.RIGHT
AND Parent.LEFT = 1 -- Given Parent Node Left Index
Wikipedia
回答6:
I found Wikipedia link has good minimized version of answer along with selected answer.
SELECT DISTINCT Child.Name
FROM ModelTable AS Child, ModelTable AS Parent
WHERE Parent.Lft < Child.Lft AND Parent.Rgt > Child.Rgt -- associate Child Nodes with ancestors
GROUP BY Child.Name
HAVING MAX(Parent.Lft) = @parentId -- Subset for those with the given Parent Node as the nearest ancestor
And, any of you try to express it with Linq, please follow the link: https://stackoverflow.com/a/25594386/361100