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无法在sqlachemy flask中设置结果对象的属性

flask-sqlalchemy sqlalchemy python

2

FlyingBurger · 技术社区 · 8 年前

我遇到了一个关于flask sqlAlchemy的问题,我可以在 放置收藏 ,但当我要为中的对象设置属性时, 地方 ,发生错误:

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join search对象的属性,有人能帮我吗?

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)

他们两个都是 类“flask”“sqlacalchemy.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)

1 回复 | 直到 8 年前

1

SuperShoot npburns224 8 年前

place_collections = Place.query.filter_by(county='BKK') 会给你一个收藏 Place 物体。这类似于 session.query(Place).filter_by(county='BKK') 简单的香草炼金术。

然而,从sqlacalchemy docs :

查询还接受ORM检测的描述符作为参数。任何时间多个类实体或基于列的实体表示为 query()函数的参数,返回结果表示为元组

关键是,当您像在这里那样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 tuples .

我的第一印象是 can't set attribute 尝试将属性值赋给 tuple ,所以我试了一下:

>>> 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。

所以我执行了一个类似于您的第二个查询(模型只是我打开的项目中的某个内容):

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

对,所以我们没有得到一个元组,我们得到一个 sqlalchemy.util._collections.result 对象。但是…

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

所以, sqlAlchemy.util.\u collections.result 是一类型属于 元组 .

如果我们试图将值赋给一个属性,而该属性不是所查询的列之一:

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

这条错误消息与我们在将属性分配给plain时得到的消息非常相似。 元组 .

那么,为什么会收到不同的错误消息呢?这个 result 对象实际上更像 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

这与您收到的错误消息相同。

所以, sqlAlchemy.util.\u collections.result 对象支持对列名称的属性访问,但由于它是 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(除非您是专门查询那些列的原因),请更改 places 查询对象:

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