SQLAlchemy教程

SQLAlchemy 删除相关对象

很容易对单个表进行删除操作。您所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有点棘手。
在我们的 sales.db 数据库中,Customer 和 Invoice 类通过一对多类型的关系映射到 customer 和 invoice 表。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")
我们设置一个会话并通过使用以下程序使用主 ID 查询来获取 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这个 x 并计算这个名称的出现次数。
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 的发票编号。上面查询的结果是2,表示相关对象没有被删除。
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 不假设删除级联;我们必须发出删除它的命令。
为了改变行为,我们在 User.addresses 关系上配置级联选项。让我们关闭正在进行的会话,使用 new declarative_base() 并重新声明 User 类,添加包括级联配置在内的地址关系。
关系函数中的cascade属性是一个以逗号分隔的级联规则列表,它决定了Session操作应该如何从父级到子级"级联"。默认情况下,它是False,这意味着它是"save-update,合并"。
可用的级联如下-
保存更新 合并 删除 删除 删除孤儿 刷新-过期
经常使用的选项是"all, delete-orphan",表示相关对象在所有情况下都应该跟随父对象,并在解除关联时删除。
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名为 Gopal Krishna 的 Customer 并查看其相关 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()
计数现在为 0,上面的脚本发出了以下 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)
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