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

一个适用于继承类的解决方案:

from itertools import chain
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()


class Mixin(object):
    def as_dict(self):
        tables = [base.__table__ for base in self.__class__.__bases__ if base not in [Base, Mixin]]
        tables.append(self.__table__)
        return {c.name: getattr(self, c.name) for c in chain.from_iterable([x.columns for x in tables])}

2016-06-08 14:11:54

其他回答

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)

2010-03-16 13:24:01

@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

使用以下SQLAlchemy代码查询数据库后:

from sqlalchemy import create_engine
from sqlalchemy.orm import sessionmaker


SQLALCHEMY_DATABASE_URL = 'sqlite:///./examples/sql_app.db'
engine = create_engine(SQLALCHEMY_DATABASE_URL, echo=True)
query = sqlalchemy.select(TABLE)
result = engine.execute(query).fetchall()

你可以使用下面的一行代码:

query_dict = [record._mapping for record in results]

2022-07-04 11:59:03

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。

2020-02-29 04:49:43

参考Alex Brasetvik的答案，你可以用一行代码来解决这个问题

row_as_dict = [dict(row) for row in resultproxy]

在Alex Brasetvik的回答的评论部分，SQLAlchemy的创建者zzzeek表示这是解决这个问题的“正确方法”。

2017-07-03 10:26:39

如何将SQLAlchemy行对象转换为Python字典?

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