如何在SQLAlchemy中查詢這些外鍵關係？

Question

我有以下3個SQL鍊金術車型

class MyModelA(db.Model): 
    a_id = db.Column(db.Integer, nullable=False, primary_key=True) 
    my_field1 = db.Column(db.String(1024), unique=True) 

class MyModelB(db.Model): 
    b_id = db.Column(db.Integer, nullable=False, primary_key=True) 
    my_field2 = db.Column(db.String(1024), nullable=True) 

    def my_method(self, arg_my_field1): 
     pass # what goes here?? 

class MyModelC(db.Model, Timestamp): 
    c_id = db.Column(db.Integer, nullable=False, primary_key=True) 
    a_id = db.Column(db.Integer, db.ForeignKey(MyModelA.a_id), default=lambda: MyModelA.query.filter(MyModelA.my_field1 == 'XYZ').one().a_id) 
    a = db.relationship('MyModelA', backref=db.backref('my_model_c')) 
    b_id = db.Column(db.Integer, db.ForeignKey(MyModelB.b_id), nullable=False) 
    b = db.relationship('MyModelB', backref=db.backref('my_model_c')) 
    my_field3 = db.Column(db.String(1024), unique=True) 
    __table_args__ = (db.UniqueConstraint('a_id', 'b_id', name='unique_constraint_aid_bid'),) 

在該方法中my_method，我想回到的MyModelC實例的my_field3字段（指向self並指向具有my_field1的情況下，MyModelA - 敏感匹配arg_my_field1）。應該有最多一個這樣的例子。如果不存在這樣的MyModelC實例，則返回None。

我應該在my_method中寫什麼SQLAlchemy查詢以實現該結果？

Answer 1

下面是答案

def my_method(self, arg_my_field1): 
    return self.my_model_c.join(MyModelA).filter(MyModelA.my_field1.ilike(arg_my_field1)).one().my_field3 

它需要lazy='dynamic'添加到MyModelC兩個關係如下圖所示：

class MyModelC(db.Model, Timestamp): 
    c_id = db.Column(db.Integer, nullable=False, primary_key=True) 
    a_id = db.Column(db.Integer, db.ForeignKey(MyModelA.a_id), default=lambda: MyModelA.query.filter(MyModelA.my_field1 == 'XYZ').one().a_id) 
    a = db.relationship('MyModelA', backref=db.backref('my_model_c', lazy='dynamic')) 
    b_id = db.Column(db.Integer, db.ForeignKey(MyModelB.b_id), nullable=False) # MODIFIED 
    b = db.relationship('MyModelB', backref=db.backref('my_model_c', lazy='dynamic')) # MODIFIED 
    my_field3 = db.Column(db.String(1024), unique=True) 
    __table_args__ = (db.UniqueConstraint('a_id', 'b_id', name='unique_constraint_aid_bid'))