Flask-SQLAlchemy 查询连接关系表

Question

我正在使用 Flask 和 SQLAlchemy 构建一个应用程序。我基本上有 3 个表：users、friendships 和 bestFriends:

一个用户可以有很多朋友，但是最好的朋友只有一个。所以我希望我的模型是相关的。 'One-to-many' 表示 'users' 和 'friendships' 之间的关系 & 'one-to-one' 表示 'users' 和 'bestFriends' 之间的关系。

这是我的模特：

from app import db
from sqlalchemy.orm import relationship, backref
from sqlalchemy import Table, Column, Integer, ForeignKey
from sqlalchemy.ext.declarative import declarative_base

class users(db.Model):

    __tablename__ = "Users"

    id = db.Column(db.Integer, primary_key=True)
    userName = db.Column(db.String, nullable=False)
    userEmail = db.Column(db.String, nullable=False)
    userPhone = db.Column(db.String, nullable=False)
    userPass = db.Column(db.String, nullable=False)

    friendsR = db.relationship('friendships', backref='friendships.friend_id', primaryjoin='users.id==friendships.user_id', lazy='joined')

    def __init__(self, userName, userEmail, userPhone, userPass):
        self.userName = userName
        self.userEmail = userEmail
        self.userPhone = userPhone
        self.userPass = userPass

    def __repr__(self):
        return '{}-{}-{}-{}'.format(self.id, self.userName, self.userEmail, self.userPhone)

class friendships(db.Model):

    __tablename__ = "Friendships"

    id = db.Column(db.Integer, primary_key=True)
    user_id = db.Column(db.Integer, ForeignKey('Users.id'), nullable=False)
    friend_id = db.Column(db.Integer, ForeignKey('Users.id'), nullable=False)

    userR = relationship('users', foreign_keys='friendships.user_id')
    friendR = relationship('users', foreign_keys='friendships.friend_id')

    def __init__(self, user_id, friend_id):
        self.user_id = user_id
        self.friend_id = friend_id


    def __repr__(self):
        return '{}-{}-{}-{}'.format(self.user_id, self.friend_id)


class bestFriends(db.Model):

    __tablename__ = "BestFriends"

    id = db.Column(db.Integer, primary_key=True)
    user_id = db.Column(db.Integer, ForeignKey('Users.id'), nullable=False)
    best_friend_id = db.Column(db.Integer, ForeignKey('Users.id'), nullable=False)

    user = relationship('users', foreign_keys='bestFriends.user_id')
    best_friend = relationship('users', foreign_keys='bestFriends.best_friend_id')


    def __init__(self, user_id, best_friend_id):

        self.user_id = user_id
        self.best_friend_id = best_friend_id


    def __repr__(self):
        return '{}-{}-{}-{}'.format(self.user_id, self.best_friend_id)

我需要能够查询登录用户的朋友列表以及该用户最好的朋友（如果存在的话）。我还需要对结果进行分页：

这是我的 app.py 函数，用于显示用户的好友：

@app.route('/friendList<int:page>', methods=['GET', 'POST'])
@app.route('/friends')
def friendList(page=1):

if not session.get('logged_in'):
        return render_template('login.html')
    else:
        userID = session['user_id']

        userList = users.query.join(friendships).add_columns(users.id, users.userName, users.userEmail, friendships.user_id, friendships.friend_id).filter(users.id == friendships.friend_id).filter(friendships.user_id == userID).paginate(page, 1, False)

        return render_template(
            'friends.html', userList=userList)

这将是代码的 Jinja 端：

{% extends "layout.html" %}
{% block body %}

<div id="pagination">
{% if userList.has_prev %}
        <a href="{{ url_for('friendList', page=userList.prev_num) }}">Back</a>
{% endif %} 

{% if userList.has_next %}
        <a href="{{ url_for('friendList', page=userList.next_num) }}">Next</a>
{% endif %}
</div>

<div style="clear:both;"></div>

<div id="innerContent">
{% if userList %}
    {% for friends in userList %}
                    <div class="contentUsers">
                        {{ friends.userName }}
                    </div>

                    <br><br><br><br>

    {% endfor %}{% else %}<div>No friends</div>
{% endif %} 

</div>
  {% endblock %}

如果我这样查询：

userList = db.session.query(users,friendships).filter(users.id == friendships.friend_id).filter(friendships.user_id == userID).paginate(page, 1, False)

我收到这个错误：

InvalidRequestError: Could not find a FROM clause to join from. Tried joining to <class 'models.friendships'>, but got: Can't determine join between 'Users' and 'Friendships'; tables have more than one foreign key constraint relationship between them. Please specify the 'onclause' of this join explicitly.

如果我这样查询：

userList = users.query.join(friendships, users.id==friendships.user_id).add_columns(users.id, users.userName, users.userEmail, friendships.user_id, friendships.friend_id).filter(users.id == friendships.friend_id).filter(friendships.user_id == userID).paginate(page, 1, False)

我收到以下错误：

TypeError: 'Pagination' object is not iterable

我仍然认为稍后的查询是正确的方法，但我认为我的 relationships/foreign 表之间的键有问题！！！

如果在 Jinja 方面，我将 .items 添加到循环中：

{% if userList.items %}
{% for friends in userList.items %}

                <div class="contentUsers">
                    {{ friends.userName }}
                </div>

                <br><br><br><br>

{% endfor %}{% else %}<div>No friends</div>

{% endif %}

根本不循环，只是显示"no friends"

的else语句

Answer 1

错误信息告诉你SQLAlchemy 无法确定如何连接users 和friendships 两个表，因为链接它们的外键不止一个。您需要明确定义连接条件。

尝试：

userList = users.query\
    .join(friendships, users.id==friendships.user_id)\
    .add_columns(users.userId, users.name, users.email, friends.userId, friendId)\
    .filter(users.id == friendships.friend_id)\
    .filter(friendships.user_id == userID)\
    .paginate(page, 1, False)

Answer 2

好的，在睡了一觉并查看 Matthewh 的建议后，我几乎找到了最终解决方案：

我的模特：

from app import db
from sqlalchemy.orm import relationship, backref
from sqlalchemy import Table, Column, Integer, ForeignKey
from sqlalchemy.ext.declarative import declarative_base

class users(db.Model):
    __tablename__ = "Users"

    id = db.Column(db.Integer, primary_key=True)
    userName = db.Column(db.String, nullable=False)
    userEmail = db.Column(db.String, nullable=False)
    userPhone = db.Column(db.String, nullable=False)
    userPass = db.Column(db.String, nullable=False)

    def __init__(self, userName, userEmail, userPhone, userPass):
        self.userName = userName
        self.userEmail = userEmail
        self.userPhone = userPhone
        self.userPass = userPass

    def __repr__(self):
        return '{}-{}-{}-{}'.format(self.id, self.userName, self.userEmail, self.userPhone)


 class friendships(db.Model):
    __tablename__ = "Friendships"

    id = db.Column(db.Integer, primary_key=True)
    user_id = db.Column(db.Integer, db.ForeignKey('Users.id'), nullable=False)
    friend_id = db.Column(db.Integer, db.ForeignKey('Users.id'), nullable=False)
    userR = db.relationship('users', foreign_keys='friendships.user_id')
    friendR = db.relationship('users', foreign_keys='friendships.friend_id')

    def __init__(self, user_id, friend_id):
        self.user_id = user_id
        self.friend_id = friend_id

    def __repr__(self):
        return '{}-{}-{}-{}'.format(self.user_id, self.friend_id)


class bestFriends(db.Model):
    __tablename__ = "BestFriends"

    id = db.Column(db.Integer, primary_key=True)
    user_id = db.Column(db.Integer, db.ForeignKey('Users.id'), nullable=False)
    best_friend_id = db.Column(db.Integer, db.ForeignKey('Users.id'), nullable=False)
    user = db.relationship('users', foreign_keys='bestFriends.user_id')
    best_friend = db.relationship('users', foreign_keys='bestFriends.best_friend_id')

    def __init__(self, user_id, best_friend_id):
        self.user_id = user_id
        self.best_friend_id = best_friend_id

    def __repr__(self):
        return '{}-{}-{}-{}'.format(self.user_id, self.best_friend_id)

我的app.py功能（显示已登录用户的好友）：

@app.route('/friendList<int:page>', methods=['GET', 'POST'])
@app.route('/friends')
  def friendList(page=1):
  if not session.get('logged_in'):
     return render_template('login.html')
  else:
     userID = session['user_id']
     userList = users.query.join(friendships, users.id==friendships.user_id).add_columns(users.id, users.userName, users.userEmail, friendships.id, friendships.user_id, friendships.friend_id).filter(friendships.friend_id == userID).paginate(page, 1, False)
     return render_template('friends.html', userList=userList)

'friends.html' 的神社方面：

{% extends "layout.html" %}

{% block body %}
    <div id="pagination">
        {% if userList.has_prev %}
            <a href="{{ url_for('friendList', page=userList.prev_num)}}">Back</a>
        {% endif %}

        {% if userList.has_next %}
            <a href="{{ url_for('friendList', page=userList.next_num)}}">Next</a>
        {% endif %}
    </div>

    <div style="clear:both;"></div>

    <div id="innerContent">
    {% if userList.items %}
        {% for friends in userList.items %}
            <div class="contentUsers">
                {{ friends.userName }}
            </div>
            <br><br><br><br>
        {% endfor %}
        {% else %}
            <div>No friends</div>
        {% endif %}
    </div>
{% endblock %}

这给了我一个像这样的对象（userList.items 中的朋友）：

(2-Carlos-carlos@carlos.com-900102030, 2, u'Carlos', u'carlos@carlos.com', 2, 2, 1)

所以我有以下 doubts/questions:

我不完全理解为查询生成的对象的结构：

-重复的用户 ID、姓名和电子邮件 -第二个名字和电子邮件

的'u'前面是什么

我没有完全理解关系模型结构：

为什么数据库模型的 'users' class 中不需要以下内容：

#friendsR = db.relationship('friendships', backref='friendships.friend_id', primaryjoin='users.id==friendships.user_id', lazy='joined')

关于标准化关系模型，我的数据库模型是否已正确定义，如此答案中所述，或者我应该如何改进？？

Flask-SQLAlchemy 查询连接关系表

Flask-SQLAlchemy Query Join relational tables

join

sqlalchemy

flask

flask-sqlalchemy