如何将SQLAlchemy行对象转换为Python字典?

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

我在这方面没有太多经验，但下面的方法似乎对我来说很有用:

dict(row)

这似乎太简单了(与这里的其他答案相比)。我错过了什么?

2016-05-18 18:21:58

其他回答

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)

2020-09-17 06:38:30

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'}

2021-03-23 19:25:54

为了完成@Anurag Uniyal的回答，这里有一个递归地遵循关系的方法:

from sqlalchemy.inspection import inspect

def to_dict(obj, with_relationships=True):
    d = {}
    for column in obj.__table__.columns:
        if with_relationships and len(column.foreign_keys) > 0:
             # Skip foreign keys
            continue
        d[column.name] = getattr(obj, column.name)

    if with_relationships:
        for relationship in inspect(type(obj)).relationships:
            val = getattr(obj, relationship.key)
            d[relationship.key] = to_dict(val) if val else None
    return d

class User(Base):
    __tablename__ = 'users'
    id = Column(Integer, primary_key=True)
    first_name = Column(TEXT)
    address_id = Column(Integer, ForeignKey('addresses.id')
    address = relationship('Address')

class Address(Base):
    __tablename__ = 'addresses'
    id = Column(Integer, primary_key=True)
    city = Column(TEXT)


user = User(first_name='Nathan', address=Address(city='Lyon'))
# Add and commit user to session to create ids

to_dict(user)
# {'id': 1, 'first_name': 'Nathan', 'address': {'city': 'Lyon'}}
to_dict(user, with_relationship=False)
# {'id': 1, 'first_name': 'Nathan', 'address_id': 1}

2019-05-23 10:45:21

行有一个_asdict()函数，它给出一个字典

In [8]: r1 = db.session.query(Topic.name).first()

In [9]: r1
Out[9]: (u'blah')

In [10]: r1.name
Out[10]: u'blah'

In [11]: r1._asdict()
Out[11]: {'name': u'blah'}

2014-01-09 19:53:55

正在迭代的表达式求值为模型对象列表，而不是行。下面是正确的用法:

for u in session.query(User).all():
    print u.id, u.name

你真的需要把它们转换成字典吗?当然，有很多方法，但是你不需要SQLAlchemy的ORM部分:

result = session.execute(User.__table__.select())
for row in result:
    print dict(row)

更新:看一下sqlalchemy.orm.attributes模块。它有一组处理对象状态的函数，这可能对您很有用，特别是instance_dict()。

2009-12-25 05:42:11

如何将SQLAlchemy行对象转换为Python字典?

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