[英]SQLAlchemy query for object with count of relationship
Lets say I have SQL Alchemy ORM classes: 可以说我有SQL Alchemy ORM类:
class Session(db.Model):
id = db.Column(db.Integer, primary_key=True)
user_agent = db.Column(db.Text, nullable=False)
class Run(db.Model):
id = db.Column(db.Integer, primary_key=True)
session_id = db.Column(db.Integer, db.ForeignKey('session.id'))
session = db.relationship('Session', backref=db.backref('runs', lazy='dynamic'))
And I want to query for essentially the following: 我想查询基本上以下内容:
((session.id, session.user_agent, session.runs.count())
for session in Session.query.order_by(Session.id.desc()))
However, this is clearly 1+n queries, which is terrible. 但是,这显然是1 + n查询,这很糟糕。 What is the correct way to do this, with 1 query?
使用1个查询执行此操作的正确方法是什么? In normal SQL, I would do this with something along the lines of:
在普通的SQL中,我会使用以下内容执行此操作:
SELECT session.id, session.user_agent, COUNT(row.id) FROM session
LEFT JOIN rows on session.id = rows.session_id
GROUP BY session.id ORDER BY session.id DESC
Construct a subquery that groups and counts session ids from runs, and join to that in your final query. 构造一个子查询,该子查询对运行中的会话ID进行分组和计数,并加入到最终查询中的会话ID。
sq = session.query(Run.session_id, func.count(Run.session_id).label('count')).group_by(Run.session_id).subquery()
result = session.query(Session, sq.c.count).join(sq, sq.c.session_id == Session.id).all()
The targeted SQL could be produced simply with: 目标SQL可以简单地生成:
db.session.query(Session, func.count(Run.id)).\
outerjoin(Run).\
group_by(Session.id).\
order_by(Session.id.desc())
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