如何使用 SQLAlchemy 将外键约束正确添加到 SQLite 数据库

Question

我是 SQLAlchemy 的新手，我正在努力弄明白。

请记住以下测试设置：

class Nine(Base):
    __tablename__ = 'nine'
    __table_args__ = (sqlalchemy.sql.schema.UniqueConstraint('nine_b', name='uq_nine_b'), )

    nine_a = sqlalchemy.Column(sqlalchemy.dialects.sqlite.INTEGER(), primary_key=True, autoincrement=False, nullable=False)
    nine_b = sqlalchemy.Column(sqlalchemy.String(20), nullable=False)


class Seven(Base):
    __tablename__ = 'seven'
    __table_args__ = (sqlalchemy.sql.schema.PrimaryKeyConstraint('seven_a', 'seven_b'),
                      sqlalchemy.sql.schema.Index('fk_seven_c_nine_a_idx', 'seven_c'),)

    seven_a = sqlalchemy.Column(sqlalchemy.dialects.sqlite.INTEGER(), nullable=False)
    seven_b = sqlalchemy.Column(sqlalchemy.dialects.sqlite.INTEGER(), nullable=False)
    seven_c = sqlalchemy.Column(sqlalchemy.dialects.sqlite.INTEGER(), sqlalchemy.ForeignKey('nine.nine_a'), nullable=False)
    seven_d = sqlalchemy.Column(sqlalchemy.dialects.sqlite.INTEGER(), nullable=False)

    nine = sqlalchemy.orm.relationship(Nine, backref=sqlalchemy.orm.backref('seven'), uselist=False)


class Three(Base):
    __tablename__ = 'three'
    __table_args__ = (sqlalchemy.sql.schema.UniqueConstraint('three_b', 'three_c', name='uq_three_b_c'),
                      sqlalchemy.sql.schema.Index('fk_three_c_seven_a_idx', 'three_c'), )

    three_a = sqlalchemy.Column(sqlalchemy.dialects.sqlite.INTEGER(), primary_key=True, autoincrement=True, nullable=False)
    three_b = sqlalchemy.Column(sqlalchemy.dialects.sqlite.INTEGER(), nullable=False)
    three_c = sqlalchemy.Column(sqlalchemy.dialects.sqlite.INTEGER(), sqlalchemy.ForeignKey('seven.seven_a'), nullable=False)

    seven = sqlalchemy.orm.relationship(Seven, backref=sqlalchemy.orm.backref('three'), uselist=False)

转换为以下 DDL：

CREATE TABLE nine (
    nine_a INTEGER NOT NULL, 
    nine_b VARCHAR(20) NOT NULL, 
    PRIMARY KEY (nine_a), 
    CONSTRAINT uq_nine_b UNIQUE (nine_b)
);

CREATE TABLE seven (
    seven_a INTEGER NOT NULL, 
    seven_b INTEGER NOT NULL, 
    seven_c INTEGER NOT NULL, 
    seven_d INTEGER NOT NULL, 
    PRIMARY KEY (seven_a, seven_b), 
    FOREIGN KEY(seven_c) REFERENCES nine (nine_a)
);

CREATE INDEX fk_seven_c_nine_a_idx ON seven (seven_c);

CREATE TABLE three (
    three_a INTEGER NOT NULL, 
    three_b INTEGER NOT NULL, 
    three_c INTEGER NOT NULL, 
    PRIMARY KEY (three_a), 
    CONSTRAINT uq_three_b_c UNIQUE (three_b, three_c), 
    FOREIGN KEY(three_c) REFERENCES seven (seven_a)
);

CREATE INDEX fk_three_c_seven_a_idx ON three (three_c);

所有桌子都是空的。然后，如下代码语句：

session.add(Nine(nine_a=1, nine_b='something'))
session.add(Nine(nine_a=2, nine_b='something else'))
session.commit()

session.add(Seven(seven_a=7, seven_b=7, seven_c=7, seven_d=7))
session.commit()

session.add(Three(three_a=3, three_b=3, three_c=3))
sessionDB.commit()

有人能解释一下为什么上面的代码片段执行没有错误吗？ FK 约束是否应该停止向 seven 或 three 中插入新行？我假设类本身对 FK 的描述有问题，但我不知道问题出在哪里（以及如何解决）。

[编辑 1]

为所有类添加 __table_args__（忘记包括它们）。

[编辑 2]

添加 DDL 以供进一步参考。

Answer 1

默认情况下，SQLite 不强制执行 ForeignKey 约束（参见此处 http://www.sqlite.org/pragma.html#pragma_foreign_keys）

要启用，请在此处遵循这些文档：http://docs.sqlalchemy.org/en/latest/dialects/sqlite.html#foreign-key-support

这是官方文档的复制粘贴：

SQLite supports FOREIGN KEY syntax when emitting CREATE statements for tables, however by default these constraints have no effect on the operation of the table.

Constraint checking on SQLite has three prerequisites:

At least version 3.6.19 of SQLite must be in use

The SQLite library must be compiled without the SQLITE_OMIT_FOREIGN_KEY or SQLITE_OMIT_TRIGGER symbols enabled.

The PRAGMA foreign_keys = ON statement must be emitted on all connections before use. SQLAlchemy allows for the PRAGMA statement to be emitted automatically for new connections through the usage of events:

from sqlalchemy.engine import Engine
from sqlalchemy import event

@event.listens_for(Engine, "connect")
def set_sqlite_pragma(dbapi_connection, connection_record):
    cursor = dbapi_connection.cursor()
    cursor.execute("PRAGMA foreign_keys=ON")
    cursor.close()

如何使用 SQLAlchemy 将外键约束正确添加到 SQLite 数据库

How to correctly add Foreign Key constraints to SQLite DB using SQLAlchemy

python

sqlite

sqlalchemy

foreign-keys

foreign-key-relationship