相关疑难解决方法(0)

我正在使用SQLite和C#,并且有一些表定义了外键.

现在,我知道默认情况下,SQLite中没有强制执行外键约束,但是我想将它们打开.

是否可以通过代码执行此操作？我查了一个相关的问题,但我不知道如何通过C#代码来完成.我正在使用Visual Studio 2008中最新的SQLite插件来设计我的表.

conn.Open();
SQLiteCommand cmd = new SQLiteCommand("PRAGMA foreign_keys = ON", conn);
cmd.ExecuteNonQuery(); 
conn.Close();

重新打开此连接时,我需要保持此更改.可能吗？

我对SQLAlchemy很新,我想弄明白.

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 = …