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

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()

其他回答

我只是花了几分钟来处理这个问题。 标记为正确的答案不尊重字段的类型。 解决方案来自于dictalchemy,添加了一些有趣的功能。 https://pythonhosted.org/dictalchemy/ 我刚刚测试过,工作正常。

Base = declarative_base(cls=DictableModel)

session.query(User).asdict()
{'id': 1, 'username': 'Gerald'}

session.query(User).asdict(exclude=['id'])
{'username': 'Gerald'}

您可以像这样将sqlalchemy对象转换为字典,并将其作为json/dictionary返回。

辅助功能:

import json
from collections import OrderedDict


def asdict(self):
    result = OrderedDict()
    for key in self.__mapper__.c.keys():
        if getattr(self, key) is not None:
            result[key] = str(getattr(self, key))
        else:
            result[key] = getattr(self, key)
    return result


def to_array(all_vendors):
    v = [ ven.asdict() for ven in all_vendors ]
    return json.dumps(v) 

驱动程序功能:

def all_products():
    all_products = Products.query.all()
    return to_array(all_products)

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。

我是一个新晋的Python程序员,遇到了使用join表获取JSON的问题。使用这里的答案中的信息,我构建了一个函数,将合理的结果返回到JSON,其中包括表名,避免使用别名或字段冲突。

简单地传递会话查询的结果:

test = Session()。查询(VMInfo、客户). join(客户).order_by (VMInfo.vm_name) .limit (50) .offset (10)

json = sqlAl2json(test)

def sqlAl2json(self, result):
    arr = []
    for rs in result.all():
        proc = []
        try:
            iterator = iter(rs)
        except TypeError:
            proc.append(rs)
        else:
            for t in rs:
                proc.append(t)

        dict = {}
        for p in proc:
            tname = type(p).__name__
            for d in dir(p):
                if d.startswith('_') | d.startswith('metadata'):
                    pass
                else:
                    key = '%s_%s' %(tname, d)
                    dict[key] = getattr(p, d)
        arr.append(dict)
    return json.dumps(arr)

有了这段代码,您还可以添加到您的查询“过滤器”或“连接”,这工作!

query = session.query(User)
def query_to_dict(query):
        def _create_dict(r):
            return {c.get('name'): getattr(r, c.get('name')) for c in query.column_descriptions}

    return [_create_dict(r) for r in query]