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

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

其他回答

class User(object):
    def to_dict(self):
        return dict([(k, getattr(self, k)) for k in self.__dict__.keys() if not k.startswith("_")])

这应该有用。

2010-02-10 14:07:02

from copy import copy

def to_record(row):
    record = copy(row.__dict__)
    del record["_sa_instance_state"]
    return record

如果不使用复制，可能会遇到错误。

2021-08-13 16:14:40

使用sqlalchemy 1.4

session.execute(select(User.id, User.username)).mappings().all()
>> [{'id': 1, 'username': 'Bob'}, {'id': 2, 'username': 'Alice'}]

2021-12-10 22:54:00

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。

2019-03-26 10:08:58

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

如何将SQLAlchemy行对象转换为Python字典?

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