Question

我在python中设置了以下SQLAlchemy模型：

class Relation(db.Model):
    __tablename__ = 'relations'

    relation_id = db.Column(
        db.Integer,
        primary_key=True,
        autoincrement=True)
    article_id = db.Column(
        db.Integer,
        db.ForeignKey("articles.id"),
        nullable=False)
    entity_id1 = db.Column(
        db.Integer,
        db.ForeignKey("entities.entity_id"),
        nullable=False)
    entity_id2 = db.Column(
        db.Integer,
        db.ForeignKey("entities.entity_id"),
        nullable=False)
    confidence = db.Column(
        db.Integer
        )


class Entity(db.Model):
    __tablename__ = 'entities'
    entity_id = db.Column(
        db.Integer,
        primary_key=True,
        autoincrement=True)
    name = db.Column(db.String(128), nullable=False)

我正在尝试将以下SQL查询字符串转换为SQLAlchemy查询（如果重要的话，请输入POSTGRES），并且主要目标是防止SQL注入：

   SELECT DISTINCT e.name, e.entity_id FROM entities e RIGHT JOIN 
   relations r ON e.entity_id=r.entity_id1 OR e.entity_id=r.entity_id2 
   WHERE (r.entity_id1 = PARAMETER OR r.entity_id2 = PARAMETER) 
   AND e.entity_id<>PARAMETER;

这是我的尝试，但是我收到的错误消息表明联接存在问题：

relations = db.session.query(Entity).join(Relation.entity_id1)\     .join(Relation.entity_id2).distinct(Entity.name).filter((Relation.entity_id1 == PARAMETER)
                                                                | (Relation.entity_id2 == PARAMTER)
                                                                & (Entity.entity_id != PARAMETER))

任何建议或调试帮助将不胜感激！

编辑：有关用例的更多信息：“实体”是无向图的节点，“关系”是边。（Relation.entity_id1 | Relation.entity_id2与Relation.entity_id2 | Relation.entity_id1相同）保证所有边缘源和目标都有相应的节点。给该目标一个entity_id = PARAMETER，找到该实体的所有相邻节点。

如何将原始SQL转换为SQLAlchemy ORM

0 个答案: