SQL-Alchemy加入查询 - 最有效的查询方式

Question

我正在尝试生成查询并难以找到在Sqlalchemy中的最有效的方法，（注意我正在使用flask-sqlalchemy

目标是找到所有用户与特定用户会面。

所以让我们说弗兰克有10个会议，我想生成一份弗兰克的所有人都会开会。

这里是我的型号：

class UserMeeting(db.Model):
    """ Associative table, links meetings to users in a many to many fashion"""
    __tablename__ = 'userMeeting'
    id = db.Column(db.Integer, primary_key = True)
    meeting_id = db.Column(db.Integer, db.ForeignKey('meeting.id'), primary_key=True)
    user_id = db.Column(db.Integer, db.ForeignKey('user.id'), primary_key=True)

class Meeting(db.Model):
    __tablename__ = "meeting"
    id = db.Column(db.Integer, primary_key = True)
    title = db.Column(db.String(128))
    #... other columns
    #associative reference
    attendees = db.relationship('UserMeeting', backref='meeting')

class User(db.Model):
    __tablename__ = 'user'
    id = db.Column(db.Integer, primary_key = True)
    email = db.Column(db.String(128), index=True, unique=True)
    password = db.Column(db.String(128))
    #associative reference
    attendingMeetings = db.relationship("UserMeeting", backref="user", cascade="all, delete-orphan") 

.

这是我尝试的原因：

#Assume frank's a user with id == 1
frank = User.query.get(1)
franks_meetings = Meeting.query.join(Meeting.attendees).filter(UserMeeting.user == frank).all()
#not efficient way of getting users in meetings with frank
users = []
for meeting in franks_meetings:
    for userMeeting in meeting.attendees:
        if userMeeting.user != frank:
            users.append(userMeeting.user)

#is there a way to just generate one query and get this data?

.

我似乎遗漏了我如何使用加入来获得此数据。任何帮助都会受到赞赏！

Answer 1

您需要使用Meeting_ID作为加入键将usermeeting表与其一起加入。您可能需要别名才能引用其两次。我不知道我是否可以在头顶上键入sqlalchemy语法，但SQL看起来像：

select distinct(b.user_id) as other_user_id
from usermeeting a
inner join usermeeting b
on a.meeting_id=b.meeting_id
where a.user_id=1 and b.user_id != 1;

.

和1是弗兰克。

哦，也得到用户详细信息。可能你最终可以直接从Sqlalchemy在Sqlalchemy中执行此操作：

select distinct(u.id), u.email
from usermeeting a
inner join usermeeting b
on a.meeting_id=b.meeting_id
inner join users u
on b.user_id=u.id
where a.user_id=1 and b.user_id != 1;

.

Answer 2

这是用于参考的查询的SQLALCHEMY版本：

#get all users in meetings with Frank, (frank.id == 1)
um = aliased(UserMeeting)
frank = User.query.get(1)

q = session.query(User).join(User.attendingMeetings).\
    filter(UserMeeting.meeting_id == um.meeting_id).\
    filter(UserMeeting.user_id != frank.id, um.user_id == frank.id)

users_meeting_with_frank = q.all()

.