I'm building a basic CMS in flask for an iPhone oriented site and I'm having a little trouble with something. I have a very small database with just 1 table (pages). Here's the model:
class Page(db.Model):
__tablename__ = 'pages'
id = db.Column(db.Integer, primary_key=True)
title = db.Column(db.String(100), nullable=False)
content = db.Column(db.Text, nullable=False)
parent_id = db.Column(db.Integer, db.ForeignKey("pages.id"), nullable=True)
As you can see, for sub pages, they just reference another page object in the parent_id field. What I'm trying to do in the admin panel is have a nested unordered list with all the pages nested in their parent pages. I have very little idea on how to do this. All i can think of is the following (which will only work (maybe—I haven't tested it) 2 levels down):
pages = Page.query.filter_by(parent_id=None)
for page in pages:
if Page.query.filter_by(parent_id=page.id):
page.sub_pages = Page.query.filter_by(parent_id=page.id)
I would then just format it into a list in the template. How would I make this work with potentially over 10 nested pages?
Thanks heaps in advance!
children = db.relationship("Page", backref=db.backref("parent", remote_side=id))
to the bottom of my Page model. and I'm looking at recursively going through everything and adding it to a tree of objects. I've probably made no sense, but that's the best way I can describe it
EDIT 2: I had a go at making a recursive function to run through all the pages and generate a big nested dictionary with all the pages and their children, but it keeps crashing python so i think it's just an infinite loop... here's the function
def get_tree(base_page, dest_dict):
dest_dict = { 'title': base_page.title, 'content': base_page.content }
children = base_page.children
if children:
dest_dict['children'] = {}
for child in children:
get_tree(base_page, dest_dict)
else:
return
and the page i'm testing it with:
@app.route('/test/')
def test():
pages = Page.query.filter_by(parent_id=None)
pages_dict = {}
for page in pages:
get_tree(page, pages_dict)
return str(pages_dict)
anyone got any ideas?
解決方案
UPD: For adjacency_list.py declarative example
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base(metadata=metadata)
class TreeNode(Base):
__tablename__ = 'tree'
id = Column(Integer, primary_key=True)
parent_id = Column(Integer, ForeignKey('tree.id'))
name = Column(String(50), nullable=False)
children = relationship('TreeNode',
# cascade deletions
cascade="all",
# many to one + adjacency list - remote_side
# is required to reference the 'remote'
# column in the join condition.
backref=backref("parent", remote_side='TreeNode.id'),
# children will be represented as a dictionary
# on the "name" attribute.
collection_class=attribute_mapped_collection('name'),
)
def __init__(self, name, parent=None):
self.name = name
self.parent = parent
def append(self, nodename):
self.children[nodename] = TreeNode(nodename, parent=self)
def __repr__(self):
return "TreeNode(name=%r, id=%r, parent_id=%r)" % (
self.name,
self.id,
self.parent_id
)
Fix recursion
def get_tree(base_page, dest_dict):
dest_dict = { 'title': base_page.title, 'content': base_page.content }
children = base_page.children
if children:
dest_dict['children'] = {}
for child in children:
get_tree(child, dest_dict)
else:
return
Use query in example for recursive fetch data from db:
# 4 level deep
node = session.query(TreeNode).\
options(joinedload_all("children", "children",
"children", "children")).\
filter(TreeNode.name=="rootnode").\
first()