Sqlalchemy 简明教程

SQLAlchemy ORM - Deleting Related Objects

对单个表执行删除操作很容易。你所要做的就是从会话中删除映射类的对象并提交该操作。但是，对多个相关表的删除操作有点棘手。

在我们的 sales.db 数据库中，Customer 和 Invoice 类被映射到客户和账单表，关系类型为一对多。我们将尝试删除 Customer 对象并查看结果。

作为快速参考，以下是 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")

我们设置了一个会话，并使用以下程序通过主键 ID 查询它来获取一个 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这个 x，并计算这个名字的出现次数。

session.delete(x)
session.query(Customer).filter_by(name = 'Gopal Krishna').count()

生成的 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 Invoice 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()

此处，10 和 14 是属于客户 Gopal Krishna 的发票号码。上述查询的结果为 2，这意味着相关的对象尚未被删除。

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 不会认为是级联删除；我们必须提供一个命令来删除它。

要更改此行为，我们配置 User.addresses 关系中的级联选项。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，并添加包含级联配置的 addresses 关系。

relationship 函数中的级联属性是一个级联规则的分隔列表，该列表确定会话操作应如何从父级级联到子级。默认情况下，它是错误的，这意味着它是“保存-更新，合并”。

可用的级联如下 −

  1. save-update

  2. merge

  3. expunge

  4. delete

  5. delete-orphan

  6. refresh-expire

通常使用的选项是“全部、删除孤立项”，以表示在所有情况下相关对象都应与父对象一起执行，并在取消关联时将其删除。

因此，重新声明的 Customer 类如下所示 −

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拥有 Gopal Krishna 名称的 Customer，并查看其关联的 Invoice 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()

使用上述脚本发出的以下 SQL，计数现在为 0 −

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