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

我对马可·马里亚尼(Marco Mariani)的回答有一个变体,以装饰者的身份表达。主要的区别是它将处理实体列表,以及安全地忽略一些其他类型的返回值(这在使用mock编写测试时非常有用):

@decorator
def to_dict(f, *args, **kwargs):
  result = f(*args, **kwargs)
  if is_iterable(result) and not is_dict(result):
    return map(asdict, result)

  return asdict(result)

def asdict(obj):
  return dict((col.name, getattr(obj, col.name))
              for col in class_mapper(obj.__class__).mapped_table.c)

def is_dict(obj):
  return isinstance(obj, dict)

def is_iterable(obj):
  return True if getattr(obj, '__iter__', False) else False

其他回答

def to_dict(row):
    return {column.name: getattr(row, row.__mapper__.get_property_by_column(column).key) for column in row.__table__.columns}


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

这个函数可能会有帮助。 当属性名与列名不同时,我找不到更好的解决方案来解决问题。

两种方式:

1.

for row in session.execute(session.query(User).statement):
    print(dict(row))

2.

selected_columns = User.__table__.columns
rows = session.query(User).with_entities(*selected_columns).all()
for row in rows :
    print(row._asdict())

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}

你可以访问SQLAlchemy对象的内部__dict__,如下所示:

for u in session.query(User).all():
    print u.__dict__
from sqlalchemy.orm import class_mapper

def asdict(obj):
    return dict((col.name, getattr(obj, col.name))
                for col in class_mapper(obj.__class__).mapped_table.c)