是否有一种简单的方法来遍历列名和值对?

我的SQLAlchemy版本是0.5.6

下面是我尝试使用dict(row)的示例代码:

import sqlalchemy
from sqlalchemy import *
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker

print "sqlalchemy version:",sqlalchemy.__version__ 

engine = create_engine('sqlite:///:memory:', echo=False)
metadata = MetaData()
users_table = Table('users', metadata,
     Column('id', Integer, primary_key=True),
     Column('name', String),
)
metadata.create_all(engine) 

class User(declarative_base()):
    __tablename__ = 'users'
    
    id = Column(Integer, primary_key=True)
    name = Column(String)
    
    def __init__(self, name):
        self.name = name

Session = sessionmaker(bind=engine)
session = Session()

user1 = User("anurag")
session.add(user1)
session.commit()

# uncommenting next line throws exception 'TypeError: 'User' object is not iterable'
#print dict(user1)
# this one also throws 'TypeError: 'User' object is not iterable'
for u in session.query(User).all():
    print dict(u)

在我的系统输出上运行这段代码:

Traceback (most recent call last):
  File "untitled-1.py", line 37, in <module>
    print dict(u)
TypeError: 'User' object is not iterable

当前回答

在大多数情况下,列名适合它们。但是你可能会像下面这样写代码:

class UserModel(BaseModel):
    user_id = Column("user_id", INT, primary_key=True)
    email = Column("user_email", STRING)

column.name“user_email”而字段名是“email”,column.name不能像以前那样工作。

sqlalchemy_base_model.py

我把答案写在这里

其他回答

如OP所述,调用dict初始化器会引发一个异常,消息为“User”对象不可迭代。所以真正的问题是如何使一个SQLAlchemy模型可迭代?

We'll have to implement the special methods __iter__ and __next__, but if we inherit directly from the declarative_base model, we would still run into the undesirable "_sa_instance_state" key. What's worse, is we would have to loop through __dict__.keys() for every call to __next__ because the keys() method returns a View -- an iterable that is not indexed. This would increase the time complexity by a factor of N, where N is the number of keys in __dict__. Generating the dict would cost O(N^2). We can do better.

我们可以实现自己的基类,它实现所需的特殊方法,并存储可以通过索引访问的列名列表,从而降低生成O(N)字典的时间复杂性。这有一个额外的好处,我们可以定义一次逻辑,并在任何时候从基类继承,我们希望我们的模型类是可迭代的。

class IterableBase(declarative_base()):
    __abstract__ = True

    def _init_keys(self):
        self._keys = [c.name for c in self.__table__.columns]
        self._dict = {c.name: getattr(self, c.name) for c in self.__table__.columns}

    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        self._init_keys()

    def __setattr__(self, name, value):
        super().__setattr__(name, value)
        if name not in ('_dict', '_keys', '_n') and '_dict' in self.__dict__:
            self._dict[name] = value

    def __iter__(self):
        self._n = 0
        return self

    def __next__(self):
        if self._n >= len(self._keys):
            raise StopIteration
        self._n += 1
        key = self._keys[self._n-1]
        return (key, self._dict[key])

现在User类可以直接从IterableBase类继承。

class User(IterableBase):
    __tablename__ = 'users'
    id = Column(Integer, primary_key=True)
    name = Column(String)

您可以确认,以User实例作为参数调用dict函数将返回所需的字典,没有"_sa_instance_state"。你可能已经注意到在IterableBase类中声明的__setattr__方法。这确保在初始化后属性发生变化或设置时更新_dict。

def main():
    user1 = User('Bob')
    print(dict(user1))
    # outputs {'id': None, 'name': 'Bob'}
    user1.id = 42
    print(dict(user1))
    # outputs {'id': 42, 'name': 'Bob'}

if __name__ == '__main__':
    main()

在python 3.8+中,我们可以使用数据类和它附带的asdict方法来实现这一点:

from dataclasses import dataclass, asdict

from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker
from sqlalchemy import Column, String, Integer, create_engine

Base = declarative_base()
engine = create_engine('sqlite:///:memory:', echo=False)


@dataclass
class User(Base):
    __tablename__ = 'users'

    id: int = Column(Integer, primary_key=True)
    name: str = Column(String)
    email = Column(String)

    def __init__(self, name):
        self.name = name
        self.email = 'hello@example.com'


Base.metadata.create_all(engine)

SessionMaker = sessionmaker(bind=engine)
session = SessionMaker()

user1 = User("anurag")
session.add(user1)
session.commit()

query_result = session.query(User).one()  # type: User
print(f'{query_result.id=:}, {query_result.name=:}, {query_result.email=:}')
# query_result.id=1, query_result.name=anurag, query_result.email=hello@example.com

query_result_dict = asdict(query_result)
print(query_result_dict)
# {'id': 1, 'name': 'anurag'}

关键是使用@dataclass装饰器,并用它的类型(name: str = column (String)行的:str部分)注释每一列。

还要注意,由于电子邮件没有注释,因此它不包括在query_result_dict中。

在@balki回答之后,从SQLAlchemy 0.8开始,您可以使用_asdict(),可用于KeyedTuple对象。这为最初的问题提供了一个非常直接的答案。只是,在你的例子中改变最后两行(for循环):

for u in session.query(User).all():
   print u._asdict()

这是因为在上面的代码中u是类型类KeyedTuple的对象,因为.all()返回KeyedTuple的列表。因此,它有_asdict()方法,该方法很好地将u作为字典返回。

WRT @STB: AFAIK的答案,任何。all()返回的是一个KeypedTuple的列表。因此,无论是否指定列,只要处理的是应用于Query对象的.all()的结果,上述方法都是有效的。

我们可以在dict中得到一个对象列表:

def queryset_to_dict(query_result):
   query_columns = query_result[0].keys()
   res = [list(ele) for ele in query_result]
   dict_list = [dict(zip(query_columns, l)) for l in res]
   return dict_list

query_result = db.session.query(LanguageMaster).all()
dictvalue=queryset_to_dict(query_result)

使用字典推导式

for u in session.query(User).all():
    print ({column.name: str(getattr(row, column.name)) for column in row.__table__.columns})