open-law/app/models/collection.py

61 lines
1.9 KiB
Python

from app import db
from app.models.utils import BaseModel
class Collection(BaseModel):
__tablename__ = "collections"
# need to redeclare id to use it in the parent relationship
id = db.Column(db.Integer, primary_key=True)
label = db.Column(db.String(256), unique=False, nullable=False)
about = db.Column(db.Text, unique=False, nullable=True)
is_root = db.Column(db.Boolean, default=False)
is_leaf = db.Column(db.Boolean, default=False)
position = db.Column(db.Integer, default=-1, nullable=True)
copy_of = db.Column(db.Integer, default=0, nullable=True)
# Foreign keys
version_id = db.Column(db.ForeignKey("book_versions.id"))
parent_id = db.Column(db.ForeignKey("collections.id"))
# Relationships
version = db.relationship("BookVersion")
children = db.relationship(
"Collection",
backref=db.backref("parent", remote_side=[id]),
viewonly=True,
order_by="asc(Collection.id)",
)
sections = db.relationship("Section")
access_groups = db.relationship(
"AccessGroup",
secondary="collections_access_groups",
) # access_groups related to current entity
def __repr__(self):
return f"<{self.id}: {self.label}>"
@property
def active_sections(self):
items = [section for section in self.sections if not section.is_deleted]
items.sort(key=lambda item: item.position)
return items
@property
def active_children(self):
items = [child for child in self.children if not child.is_deleted]
items.sort(key=lambda item: item.position)
return items
@property
def book_id(self):
return self.version.book_id
@property
def sub_collections(self):
return [
sub_collection
for sub_collection in self.children
if not sub_collection.is_deleted
]