SQLAlchemy 仅查找包含特定子代的父代答案

【问题标题】：SQLAlchemy find only parents that contain specific childrenSQLAlchemy 仅查找包含特定子代的父代
【发布时间】：2020-01-27 20:07:56
【问题描述】：

我有以下多对多模型：


class Association(db.Model):
    parent_id = db.Column(db.Integer, db.ForeignKey(
        'parent.id'), primary_key=True)
    child_id = db.Column(db.Integer, db.ForeignKey(
        'child.id'), primary_key=True)
    child = db.relationship("Child", back_populates="parents")
    parent = db.relationship("Parent", back_populates="children")


class Parent(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    children = db.relationship("Association", back_populates='parent', 
                               lazy='dynamic', cascade="save-update, merge, delete, delete-orphan")

class Child(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    parents = db.relationship("Association", back_populates='child', 
                              lazy='dynamic', cascade="save-update, merge, delete, delete-orphan")

如何获取列表中具有确切子 ID 的所有父母（例如：[5,6]）并且没有其他 ID？

【问题讨论】：

检查来自red-gate.com/simple-talk/sql/t-sql-programming/…的“精确除法”

标签： python sqlalchemy many-to-many flask-sqlalchemy

【解决方案1】：

如果您已确定子查询中的子 ID，则可以从关联表到该子查询进行左连接并检查计数是否正确：

from sqlalchemy import func, and_

Values = [1 ,3]
childcount = len(Values)
childsquery = db.session.query(Child).filter(Child.id.in_(Values)).subquery()             
resultparents = db.session.query(Association.parent_id).outerjoin(childsquery).group_by(Association.parent_id).having(and_(func.count(Association.parent_id) == childcount, func.count(childsquery.c.id) ==  childcount))

备注：我认为你的模型需要修正：

class Parent(db.Model):
   id = db.Column(db.Integer, primary_key=True)
   children = db.relationship("Association", back_populates='parent', 
            lazy='dynamic', cascade="save-update, merge, delete, delete-orphan")

class Child(db.Model):
   id = db.Column(db.Integer, primary_key=True)
   parents = db.relationship("Association", back_populates='child', 
                          lazy='dynamic', cascade="save-update, merge, delete, delete orphan")

【讨论】：

【解决方案2】：

我最终使用以下代码解决了它：

ids = [c1, c2, c3]
q1 = Parent.query.filter(Parent.children.any(
Association.child_id == c1)).filter(Parent.children.any(Association.child_id == c2))
.filter(Parent.children.any(Association.child_id == c3))
.filter(~Parent.children.any(Association.child_id.notin_(ids)))
.all()

但我很确定一定有更好的方法来做到这一点。

【讨论】：