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[英]Error while inseting flask SQLAlchemy Inserting Data in a Many-to-Many Relationship with Association Table
[英]SQLAlchemy Inserting Data in a Many-to-Many Relationship with Association Table
我已經看到了一些與此類似的問題,但是沒有一個問題在頭上很明顯。 基本上我在使用SQLAlchemy的Flask應用程序中有三個表模型Center()
, Business()
和CenterBusiness()
。 目前我正在以這種方式添加到所述關系中:
biz = Business(typId=form.type.data, name=form.name.data,
contact=form.contact.data, phone=form.phone.data)
db.session.add(biz)
db.session.commit()
assoc = CenterBusiness(bizId=biz.id, cenId=session['center'])
db.session.add(assoc)
db.session.commit()
正如你所看到的那樣有點難看,我知道有一種方法可以在一次點擊時按照它們的定義來實現。 我在SQLAlchemy的文檔中看到他們對使用這樣的表有一個解釋,但我似乎無法讓它工作。
#Directly from SQLAlchemy Docs
p = Parent()
a = Association(extra_data="some data")
a.child = Child()
p.children.append(a)
#My Version Using my Tables
center = Center.query.get(session['center']
assoc = CenterBusiness()
assoc.business = Business(typId=form.type.data, name=form.name.data,
contact=form.contact.data, phone=form.phone.data)
center.businesses.append(assoc)
db.session.commit()
不幸的是,這似乎沒有做到這一點......任何幫助將不勝感激,下面我發布了所涉及的模型。
class Center(db.Model):
id = db.Column(MEDIUMINT(8, unsigned=True), primary_key=True,
autoincrement=False)
phone = db.Column(VARCHAR(10), nullable=False)
location = db.Column(VARCHAR(255), nullable=False)
businesses = db.relationship('CenterBusiness', lazy='dynamic')
employees = db.relationship('CenterEmployee', lazy='dynamic')
class Business(db.Model):
id = db.Column(MEDIUMINT(8, unsigned=True), primary_key=True,
autoincrement=True)
typId = db.Column(TINYINT(2, unsigned=True),
db.ForeignKey('biz_type.id',
onupdate='RESTRICT',
ondelete='RESTRICT'),
nullable=False)
type = db.relationship('BizType', backref='businesses',
lazy='subquery')
name = db.Column(VARCHAR(255), nullable=False)
contact = db.Column(VARCHAR(255), nullable=False)
phone = db.Column(VARCHAR(10), nullable=False)
documents = db.relationship('Document', backref='business',
lazy='dynamic')
class CenterBusiness(db.Model):
cenId = db.Column(MEDIUMINT(8, unsigned=True),
db.ForeignKey('center.id',
onupdate='RESTRICT',
ondelete='RESTRICT'),
primary_key=True)
bizId = db.Column(MEDIUMINT(8, unsigned=True),
db.ForeignKey('business.id',
onupdate='RESTRICT',
ondelete='RESTRICT'),
primary_key=True)
info = db.relationship('Business', backref='centers',
lazy='joined')
archived = db.Column(TINYINT(1, unsigned=True), nullable=False,
server_default='0')
我能夠使這個工作,我的問題在下面的代碼中出現(粗體錯誤):
#My Version Using my Tables
center = Center.query.get(session['center']
assoc = CenterBusiness()
**assoc.info** = Business(typId=form.type.data, name=form.name.data,
contact=form.contact.data, phone=form.phone.data)
center.businesses.append(assoc)
db.session.commit()
正如我在問題中的評論所解釋的那樣:
好吧我的問題是我沒有使用我在CenterBusiness模型中的關系鍵“info”來定義附加的關聯。 我說的是center.business認為在這種情況下商業這個詞是任意的。 但是,我需要實際引用這種關系。 因此,我在CenterBusiness中設置的相應密鑰就是信息。
我仍然會接受任何更新和/或更好的方法來處理這種情況,雖然我認為這是當時最好的路線。
下面的例子可以幫助你更多細節http://docs.sqlalchemy.org/en/latest/orm/extensions/associationproxy.html
class User(Base):
__tablename__ = 'user'
id = Column(Integer, primary_key=True)
name = Column(String(64))
# association proxy of "user_keywords" collection
# to "keyword" attribute
keywords = association_proxy('user_keywords', 'keyword')
def __init__(self, name):
self.name = name
class UserKeyword(Base):
__tablename__ = 'user_keyword'
user_id = Column(Integer, ForeignKey('user.id'), primary_key=True)
keyword_id = Column(Integer, ForeignKey('keyword.id'), primary_key=True)
special_key = Column(String(50))
# bidirectional attribute/collection of "user"/"user_keywords"
user = relationship(User,
backref=backref("user_keywords",
cascade="all, delete-orphan")
)
# reference to the "Keyword" object
keyword = relationship("Keyword")
def __init__(self, keyword=None, user=None, special_key=None):
self.user = user
self.keyword = keyword
self.special_key = special_key
class Keyword(Base):
__tablename__ = 'keyword'
id = Column(Integer, primary_key=True)
keyword = Column('keyword', String(64))
def __init__(self, keyword):
self.keyword = keyword
def __repr__(self):
return 'Keyword(%s)' % repr(self.keyword)
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