我有一个自引用的邻接列表关系,我创建了以下SqlAlchemy文档。该模型看起来像:
class Menu(db.Model):
id = db.Column(db.Integer, primary_key = True)
title = db.Column(db.String(255))
ordering = db.Column(db.SmallInteger, default = '1')
parent_id = db.Column(db.Integer, db.ForeignKey('menu.id'))
children = db.relationship("Menu",
cascade="all, delete-orphan",
backref = db.backref('parent', remote_side=[id]),
collection_class = attribute_mapped_collection('id'))
我真正想要的是对该模型进行查询并获取如下数据:
root --+---> child1
+---> child2 --+--> subchild1
| +--> subchild2--+--> and so on,
| +--> and so on, if exists
+---> child3 --+--> subchild1
+--> ...
+--> ...
+--> ...
将用以下数据表示:
id parent_id title
--- ------- ----
1 NULL root
2 1 child1
3 1 child2
4 3 subchild1
5 3 subchild2
7 5 so_on1
8 5 so_on2
6 1 child3
9 6 subchild1
如何查询以检索上述表示的数据?
答案 0 :(得分:6)
您可以使用resucrsive=True
获取具有常量查询数的菜单项树,这将转换为SQL WITH RECURSIVE
(在某些RDBMS上不可用)。如果树很大并且您想节省一些时间,这可能很有用。
# db - SQLA session
# pick a root of the menu tree
root = Menu.query.filter(Menu.parent_id == None).first()
# get ids of all menu items in tree with recursive query
included = db.query(
Menu.id
).filter(
Menu.parent_id == root.id
).cte(name="included", recursive=True)
included_alias = aliased(included, name="parent")
menu_alias = aliased(Menu, name="child")
included = included.union_all(
db.query(
menu_alias.id
).filter(
menu_alias.parent_id == included_alias.c.id
)
)
# include the root's id and extract ids from tuples
menu_ids = map(
lambda _tuple: _tuple[0],
[(root.id,)] + db.query(included.c.id).distinct().all(),
)
# fetch SQLA models
menus = Menu.query.filter(Menu.id.in_(menu_ids)).all()
如果您不需要速度,只需使用children
关系在菜单项上运行DFS或BFS。