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

你可以试着这样做。

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

它使用查询对象中的内置方法返回查询对象的字典对象。

引用:https://docs.sqlalchemy.org/en/latest/orm/query.html

2019-01-21 04:30:38

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

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

2015-07-10 00:46:15

@zzzeek在评论中写道:

注意，这是现代版本的正确答案 SQLAlchemy，假设“row”是核心行对象，而不是orm映射对象实例。

for row in resultproxy:
    row_as_dict = row._mapping  # SQLAlchemy 1.4 and greater
    # row_as_dict = dict(row)  # SQLAlchemy 1.3 and earlier

行背景。_mapping, SQLAlchemy 1.4新增:https://docs.sqlalchemy.org/en/stable/core/connections.html#sqlalchemy.engine.Row._mapping

2009-12-24 12:45:04

两种方式:

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

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

2018-02-06 23:57:47

如何将SQLAlchemy行对象转换为Python字典?

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