解决方法
文章
Managing Hierarchical Data in MySQL给出了如何使用嵌套集的一个很好的例子,并给出了许多常见查询的示例,包括这一个.
这是如何找到节点的直接子节点：
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 = '**[[MY NODE]]**'
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;
然后将它与叶子节点的rgt等于lft 1这一事实相结合,然后进行设置.原谅双关语.
