我有一张桌子,是我自己做的.我的样本表是这样的
-- Table genealogy
+--------+---------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+--------+---------+------+-----+---------+-------+
| parent | int(11) | NO | PRI | NULL | |
| child | int(11) | NO | PRI | NULL | |
+--------+---------+------+-----+---------+-------+
2 rows in set (0.01 sec)这个概念是这样的(做好准备):
为了找到用户的孩子,我看用户的第一层,寻找用户的外孙,寻找用户的孙子,寻找用户的子代,找出第三层的子代,寻找parent=1.
它可能会有点混乱,但我的查询工作得很好。我想知道是否有一种更快/更优化的方法,因为我只是在各地重复我的sql。
-- Get Child
SELECT parent, child AS '1st Level' FROM genealogy WHERE parent=1;
-- Get Grandchild
SELECT a.parent, b.child AS '2nd Level' FROM (SELECT * FROM genealogy WHERE parent=1) a INNER JOIN genealogy b ON a.child=b.parent;
-- Get Great-Grandchild
SELECT a.parent, b.child AS '3rd Level' FROM (SELECT a.parent, b.child FROM (SELECT * FROM genealogy WHERE parent=1) a INNER JOIN genealogy b ON a.child=b.parent) a INNER JOIN genealogy b ON a.child=b.parent;
-- And the list goes on发布于 2011-11-28 15:42:20
您可以使用另一个树结构(称为嵌套集):
-- Table genealogy
+--------+---------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+--------+---------+------+-----+---------+-------+
| id | int(11) | NO | PRI | NULL | |
| lft | int(11) | NO | NO | NULL | |
| rgt | int(11) | NO | NO | NULL | |
+--------+---------+------+-----+---------+-------+这将使您能够选择树内一个节点的所有子程序(以及大子节点和大子节点等等)。
例如,要选择所有家谱id 1的子代:
SELECT g2.*
FROM genealogy AS g1, genealogy AS g2
WHERE
(g1.lft BETWEEN g2.lft AND g2.rgt)
AND
(g1.id = 1);有关更多信息和更多示例,您可能会查看以下内容:http://www.ibase.ru/devinfo/DBMSTrees/sqltrees.html
https://stackoverflow.com/questions/8298131
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