Django可以很好地自动序列化从DB返回到JSON格式的ORM模型。

如何序列化SQLAlchemy查询结果为JSON格式?

我试过jsonpickle。编码,但它编码查询对象本身。 我尝试了json.dumps(items),但它返回

TypeError: <Product('3', 'some name', 'some desc')> is not JSON serializable

将SQLAlchemy ORM对象序列化为JSON /XML真的那么难吗?它没有任何默认序列化器吗?现在序列化ORM查询结果是非常常见的任务。

我所需要的只是返回SQLAlchemy查询结果的JSON或XML数据表示。

需要在javascript datagird中使用JSON/XML格式的SQLAlchemy对象查询结果(JQGrid http://www.trirand.com/blog/)


当前回答

当使用sqlalchemy连接到db I时,这是一个高度可配置的简单解决方案。使用熊猫。

import pandas as pd
import sqlalchemy

#sqlalchemy engine configuration
engine = sqlalchemy.create_engine....

def my_function():
  #read in from sql directly into a pandas dataframe
  #check the pandas documentation for additional config options
  sql_DF = pd.read_sql_table("table_name", con=engine)

  # "orient" is optional here but allows you to specify the json formatting you require
  sql_json = sql_DF.to_json(orient="index")

  return sql_json

其他回答

向任何模型添加一个_dict方法的动态方法

from sqlalchemy.inspection import inspect

def implement_as_dict(model):
    if not hasattr(model,"as_dict"):
        column_names=[]
        imodel = inspect(model)
        for c in imodel.columns:
            column_names.append(c.key)

        #define model.as_dict()
        def as_dict(self):
            d = {}
            for c in column_names:
                d[c] = getattr(self,c)
            return d

        setattr(model,"as_dict",as_dict)

#model definition
class User(Base):
    __tablename__ = 'users'

    id = Column(Integer, primary_key=True)
    name = Column(String)
# adding as_dict definition to model
implement_as_dict(User)

然后你可以使用

user = session.query(User).filter_by(name='rick').first() 

user.as_dict()
#sample output 
{"id":1,"name":"rick"}

在Flask下,它工作并处理datatime字段,转换类型字段 “时间”:datetime。Datetime(2018, 3, 22, 15, 40)成 “时间”:“2018-03-22 15:40:00”:

obj = {c.name: str(getattr(self, c.name)) for c in self.__table__.columns}

# This to get the JSON body
return json.dumps(obj)

# Or this to get a response object
return jsonify(obj)

2023年末

我的实现

def obj_to_dict(obj, remove=['_sa_instance_state'], debug=False):
    result = {}

    if type(obj).__name__ == "Row":
        return dict(obj)

    obj = obj.__dict__
    for key in obj:
        if key in remove:
            continue

        result[key] = obj[key]

    if debug:
        print(result)

    return result

我已经成功地使用了这个包:https://github.com/n0nSmoker/SQLAlchemy-serializer

你可以在模型上这样做:

from sqlalchemy_serializer import SerializerMixin

class SomeModel(db.Model, SerializerMixin):
    ...

它添加了完全递归的to_dict:

item = SomeModel.query.filter(...).one()
result = item.to_dict()

它可以让你制定规则来避免无限递归:

result = item.to_dict(rules=('-somefield', '-some_relation.nested_one.another_nested_one'))

虽然最初的问题可以追溯到很久以前,但这里的答案数量(以及我自己的经验)表明,这是一个不平凡的问题,有许多不同的方法,不同的复杂性和不同的权衡。

这就是为什么我构建了SQLAthanor库,它扩展了SQLAlchemy的声明性ORM,支持可配置的序列化/反序列化,您可能想看看。

该库支持:

Python 2.7, 3.4, 3.5, and 3.6. SQLAlchemy versions 0.9 and higher serialization/de-serialization to/from JSON, CSV, YAML, and Python dict serialization/de-serialization of columns/attributes, relationships, hybrid properties, and association proxies enabling and disabling of serialization for particular formats and columns/relationships/attributes (e.g. you want to support an inbound password value, but never include an outbound one) pre-serialization and post-deserialization value processing (for validation or type coercion) a pretty straightforward syntax that is both Pythonic and seamlessly consistent with SQLAlchemy's own approach

你可以在这里查看(我希望!)全面的文档:https://sqlathanor.readthedocs.io/en/latest

希望这能有所帮助!