📅 最后修改于: 2020-11-27 07:54:27 🧑 作者: Mango
在单个表上执行删除操作很容易。您要做的就是从会话中删除映射类的对象并提交操作。但是，对多个相关表的删除操作并不复杂。
在我们的sales.db数据库中，Customer和Invoice类通过一对多关系映射到customer和invoice表。我们将尝试删除客户对象并查看结果。
作为快速参考，下面是Customer和Invoice类的定义-
from sqlalchemy import create_engine, ForeignKey, Column, Integer, String
engine = create_engine('sqlite:///sales.db', echo = True)
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
from sqlalchemy.orm import relationship
class Customer(Base):
__tablename__ = 'customers'
id = Column(Integer, primary_key = True)
name = Column(String)
address = Column(String)
email = Column(String)
class Invoice(Base):
__tablename__ = 'invoices'
id = Column(Integer, primary_key = True)
custid = Column(Integer, ForeignKey('customers.id'))
invno = Column(Integer)
amount = Column(Integer)
customer = relationship("Customer", back_populates = "invoices")
Customer.invoices = relationship("Invoice", order_by = Invoice.id, back_populates = "customer")
我们使用以下程序通过使用主要ID查询会话来设置会话并获取Customer对象-
from sqlalchemy.orm import sessionmaker
Session = sessionmaker(bind=engine)
session = Session()
x = session.query(Customer).get(2)
在我们的示例表中，x.name恰好是“ Gopal Krishna”。让我们从会话中删除此x并计算该名称的出现。
session.delete(x)
session.query(Customer).filter_by(name = 'Gopal Krishna').count()
结果SQL表达式将返回0。
SELECT count(*)
AS count_1
FROM (
SELECT customers.id
AS customers_id, customers.name
AS customers_name, customers.address
AS customers_address, customers.email
AS customers_email
FROM customers
WHERE customers.name = ?)
AS anon_1('Gopal Krishna',) 0
但是，x的相关发票对象仍然存在。可以通过以下代码进行验证-
session.query(Invoice).filter(Invoice.invno.in_([10,14])).count()
在这里，10和14是属于客户Gopal Krishna的发票编号。以上查询结果为2，表示相关对象尚未删除。
SELECT count(*)
AS count_1
FROM (
SELECT invoices.id
AS invoices_id, invoices.custid
AS invoices_custid, invoices.invno
AS invoices_invno, invoices.amount
AS invoices_amount
FROM invoices
WHERE invoices.invno IN (?, ?))
AS anon_1(10, 14) 2
这是因为SQLAlchemy不假定删除层叠。我们必须给出一个删除它的命令。
要更改行为，我们在User.addresses关系上配置了层叠选项。让我们关闭正在进行的会话，使用新的declarative_base()并重新声明User类，添加包括级联配置在内的地址关系。
关系函数的层叠属性是用逗号分隔的层叠规则列表，该列表确定如何将会话操作从父级“级联”到子级。默认情况下，它为False，这意味着它是“保存更新，合并”。
可用的级联如下-
通常使用的选项是“ all，delete-orphan”，表示在所有情况下相关对象应与父对象一起跟随，并在取消关联时被删除。
因此，重新声明的客户类别如下所示-
class Customer(Base):
__tablename__ = 'customers'
id = Column(Integer, primary_key = True)
name = Column(String)
address = Column(String)
email = Column(String)
invoices = relationship(
"Invoice",
order_by = Invoice.id,
back_populates = "customer",
cascade = "all,
delete, delete-orphan"
)
让我们使用以下程序删除具有Gopal Krishna名称的Customer，并查看其相关发票对象的数量-
from sqlalchemy.orm import sessionmaker
Session = sessionmaker(bind = engine)
session = Session()
x = session.query(Customer).get(2)
session.delete(x)
session.query(Customer).filter_by(name = 'Gopal Krishna').count()
session.query(Invoice).filter(Invoice.invno.in_([10,14])).count()
计数现在为0，上面的脚本发出以下SQL-
SELECT customers.id
AS customers_id, customers.name
AS customers_name, customers.address
AS customers_address, customers.email
AS customers_email
FROM customers
WHERE customers.id = ?
(2,)
SELECT invoices.id
AS invoices_id, invoices.custid
AS invoices_custid, invoices.invno
AS invoices_invno, invoices.amount
AS invoices_amount
FROM invoices
WHERE ? = invoices.custid
ORDER BY invoices.id (2,)
DELETE FROM invoices
WHERE invoices.id = ? ((1,), (2,))
DELETE FROM customers
WHERE customers.id = ? (2,)
SELECT count(*)
AS count_1
FROM (
SELECT customers.id
AS customers_id, customers.name
AS customers_name, customers.address
AS customers_address, customers.email
AS customers_email
FROM customers
WHERE customers.name = ?)
AS anon_1('Gopal Krishna',)
SELECT count(*)
AS count_1
FROM (
SELECT invoices.id
AS invoices_id, invoices.custid
AS invoices_custid, invoices.invno
AS invoices_invno, invoices.amount
AS invoices_amount
FROM invoices
WHERE invoices.invno IN (?, ?))
AS anon_1(10, 14)
0