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我有两个模型 -

车站:

class Station(db.Model):
    """Measurement station"""
    __tablename__ = "station"

    reference = db.Column(db.Text, nullable=False, primary_key=True)
    name = db.Column(db.Text, nullable=False, primary_key=True)
    status = db.Column(db.Text)


    river_analyses = db.relationship("RiverAnalysis")

河流分析:

class RiverAnalysis(db.Model):
    __tablename__ = "river_analysis"
    id = db.Column(db.Integer, nullable=False, primary_key=True)
    reference = db.Column(db.Text, db.ForeignKey("station.reference"), nullable=False)

    station_ref = db.relationship("Station", foreign_keys=reference)

当我尝试flask db upgrade使用此迁移代码运行时

def upgrade():
    # ### commands auto generated by Alembic - please adjust! ###
    op.add_column('river_analysis', sa.Column('reference', sa.Text(), nullable=False))
    op.create_foreign_key(None, 'river_analysis', 'station', ['reference'], ['reference'])

我得到这个错误:(psycopg2.ProgrammingError) there is no unique constraint matching given keys for referenced table "station"

为什么我得到这个即使station.reference是主键Station

感谢您的任何帮助,您可以提供。

4

1 回答 1

1

reference单独不是你的主键,而是一对(reference, name)。为了引用该键,您需要一个复合外键约束

class RiverAnalysis(db.Model):
    __tablename__ = "river_analysis"
    id = db.Column(db.Integer, nullable=False, primary_key=True)
    reference = db.Column(db.Text, nullable=False)
    name = db.Column(db.Text, nullable=False)

    __table_args__ = (
        db.ForeignKeyConstraint(
            [reference, name],
            ["station.reference", "station.name"]),
    )
于 2019-04-07T18:20:24.813 回答