python – 无法在SQLAlchemy flask中设置结果对象的属性

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我遇到了Flask-SQLAlchemy的问题,我可以在place_collections中设置对象的属性,但是当我想在场所中设置对象的属性时,发生了一个错误：

Traceback (most recent call last):
  File "/Users/user/PycharmProjects/website/venv/lib/python3.6/site-packages/flask/app.py", line 2309, in __call__
    return self.wsgi_app(environ, start_response)
  File "/Users/user/PycharmProjects/website/venv/lib/python3.6/site-packages/werkzeug/contrib/fixers.py", line 152, in __call__
    return self.app(environ, start_response)
  File "/Users/user/PycharmProjects/website/venv/lib/python3.6/site-packages/flask/app.py", line 2295, in wsgi_app
    response = self.handle_exception(e)
  File "/Users/user/PycharmProjects/website/venv/lib/python3.6/site-packages/flask/app.py", line 1741, in handle_exception
    reraise(exc_type, exc_value, tb)
  File "/Users/user/PycharmProjects/website/venv/lib/python3.6/site-packages/flask/_compat.py", line 35, in reraise
    raise value
  File "/Users/user/PycharmProjects/website/venv/lib/python3.6/site-packages/flask/app.py", line 2292, in wsgi_app
    response = self.full_dispatch_request()
  File "/Users/user/PycharmProjects/website/venv/lib/python3.6/site-packages/flask/app.py", line 1815, in full_dispatch_request
    rv = self.handle_user_exception(e)
  File "/Users/user/PycharmProjects/website/venv/lib/python3.6/site-packages/flask/app.py", line 1718, in handle_user_exception
    reraise(exc_type, exc_value, tb)
  File "/Users/user/PycharmProjects/website/venv/lib/python3.6/site-packages/flask/_compat.py", line 35, in reraise
    raise value
  File "/Users/user/PycharmProjects/website/venv/lib/python3.6/site-packages/flask/app.py", line 1813, in full_dispatch_request
    rv = self.dispatch_request()
  File "/Users/user/PycharmProjects/website/venv/lib/python3.6/site-packages/flask/app.py", line 1799, in dispatch_request
    return self.view_functions[rule.endpoint](**req.view_args)
  File "/Users/user/Document/Python/Test/Code/app/main/views.py", line 108, in index
    place.distance = round(distance_computing, 1)
AttributeError: can't set attribute

如何设置连接搜索对象的属性,有人可以帮助我吗？

place_collections = Place.query.filter_by(county='BKK')

places = db.session.query(Place.roll_number, Place.name, Place.website, Place.address, Place.distance)
            .outerjoin(Rank, Rank.place_id == Place.place_id)

它们都是类’flask_sqlalchemy.BaseQuery’类型.

for place in place_collections:
       distance_computing = Place.distance_calculator(float(place.lat), float(place.lng),
                                                            data['lat'], data['lng'])

       place.distance = round(distance_computing, 1)

for place in places:
       distance_computing = Place.distance_calculator(float(school.lat), float(school.lng),
                                                            data['lat'], data['lng'])

       place.distance = round(distance_computing, 1)

模型

class Rank(db.Model):
    __tablename__ = 'rank'
    place_id = db.Column(db.String(50))
    name = db.Column(db.String(255), nullable=False, primary_key=True)
    rank = db.Column(db.Integer)

class Place(db.Model):
    _tablename_ = 'school'
    place_id = db.Column(db.String(50), primary_key=True)
    roll_number = db.Column(db.String(50))
    name = db.Column(db.String(255), nullable=False)
    address = db.Column(db.String(255))
    distance = db.Column(db.String(50))
    rank = db.relationship('Rank',backref=db.backref('roll_number1'), lazy=True, uselist=False)

解决方法:

place_collections = Place.query.filter_by(county =’BKK’)将返回Place对象的集合.这类似于普通的SQLAlchemy中的session.query(Place).filter_by(county =’BKK’).

但是,从SQLAlchemy docs：

The Query also accepts ORM-instrumented descriptors as arguments. Any
time multiple class entities or column-based entities are expressed as
arguments to the query() function, the return result is expressed as
tuples

关键点在于,当您在此处指定要查询的特定列时：

places = db.session.query(Place.roll_number, Place.name,
    Place.website, Place.address, Place.distance).\
    outerjoin(Rank, Rank.place_id == Place.place_id)

结果表示为元组的集合.

我的第一印象是,无法设置属性是一个奇怪的错误消息,从尝试将属性值分配给元组接收,所以我尝试了：

>>> t = tuple()
>>> t.attribute = 9
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'tuple' object has no attribute 'attribute'

这不是您收到的错误消息.

所以我执行了类似于你的第二个查询(模型只是我打开的项目中的内容)：

>>> q = session.query(Racecard.id, Racecard.meeting)
>>> a_result = q[0]
>>> type(a_result)
<class 'sqlalchemy.util._collections.result'>

是的,所以我们没有得到元组,我们得到一个sqlalchemy.util._collections.result对象.但..

>>> issubclass(type(a_result), tuple)
True

因此,sqlalchemy.util._collections.result是一种元组.

如果我们尝试为不是查询的列之一的属性赋值：

>>> a_result.newattribute = 'something'
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'result' object has no attribute 'newattribute'

该错误消息非常类似于我们在将属性分配给普通元组时所获得的消息.

那你为什么得到一个不同的错误信息呢？结果对象实际上更像是一个namedtuple：

>>> from collections import namedtuple
>>> Racecard = namedtuple('Racecard', 'id, meeting')
>>> rc = Racecard(1, 'Doomben')
>>> rc.meeting = 'Eagle Farm'
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: can't set attribute

您收到的错误消息相同.

因此,sqlalchemy.util._collections.result对象支持列名的属性访问,但由于它是一个元组,它仍然是不可变的,因此您无法写入这些属性.

要修复您的特定错误(除非有某些原因您只是专门查询这些列),请将您的地点查询更改为：

places = db.session.query(Place).outerjoin(Rank, Rank.place_id == Place.place_id)

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