相关疑难解决方法(0)

我可以使用以下代码删除表,但不知道如何使用约束执行相同的操作:

IF EXISTS(SELECT 1 FROM sys.objects WHERE OBJECT_ID = OBJECT_ID(N'TableName') AND type = (N'U')) DROP TABLE TableName
go 

我还使用此代码添加约束:

ALTER TABLE [dbo].[TableName] 
  WITH CHECK ADD CONSTRAINT [FK_TableName_TableName2] FOREIGN KEY([FK_Name])
    REFERENCES [dbo].[TableName2] ([ID])
go

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 = …