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

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)

其他回答

你可以像这样使用SqlAlchemy的自省:

mysql = SQLAlchemy()
from sqlalchemy import inspect

class Contacts(mysql.Model):  
    __tablename__ = 'CONTACTS'
    id = mysql.Column(mysql.Integer, primary_key=True)
    first_name = mysql.Column(mysql.String(128), nullable=False)
    last_name = mysql.Column(mysql.String(128), nullable=False)
    phone = mysql.Column(mysql.String(128), nullable=False)
    email = mysql.Column(mysql.String(128), nullable=False)
    street = mysql.Column(mysql.String(128), nullable=False)
    zip_code = mysql.Column(mysql.String(128), nullable=False)
    city = mysql.Column(mysql.String(128), nullable=False)
    def toDict(self):
        return { c.key: getattr(self, c.key) for c in inspect(self).mapper.column_attrs }

@app.route('/contacts',methods=['GET'])
def getContacts():
    contacts = Contacts.query.all()
    contactsArr = []
    for contact in contacts:
        contactsArr.append(contact.toDict()) 
    return jsonify(contactsArr)

@app.route('/contacts/<int:id>',methods=['GET'])
def getContact(id):
    contact = Contacts.query.get(id)
    return jsonify(contact.toDict())

从下面的答案中得到启发: 将sqlalchemy行对象转换为python dict

也许你可以使用这样的类

from sqlalchemy.ext.declarative import declared_attr
from sqlalchemy import Table


class Custom:
    """Some custom logic here!"""

    __table__: Table  # def for mypy

    @declared_attr
    def __tablename__(cls):  # pylint: disable=no-self-argument
        return cls.__name__  # pylint: disable= no-member

    def to_dict(self) -> Dict[str, Any]:
        """Serializes only column data."""
        return {c.name: getattr(self, c.name) for c in self.__table__.columns}

Base = declarative_base(cls=Custom)

class MyOwnTable(Base):
    #COLUMNS!

所有对象都有to_dict方法

下面的代码将sqlalchemy结果序列化为json。

import json
from collections import OrderedDict


def asdict(self):
    result = OrderedDict()
    for key in self.__mapper__.c.keys():
        if getattr(self, key) is not None:
            result[key] = str(getattr(self, key))
        else:
            result[key] = getattr(self, key)
    return result


def to_array(all_vendors):
    v = [ ven.asdict() for ven in all_vendors ]
    return json.dumps(v) 

叫有趣,

def all_products():
    all_products = Products.query.all()
    return to_array(all_products)

https://flask-restplus.readthedocs.io/en/stable/marshalling.html

from flask_restplus import fields, Namespace, marshal
api = Namespace("Student data")
db_data = Student_details.query.all()
data_marshal_obj = api.model(" Data", {
    "id": fields.String(),
    "number": fields.Integer(),
    "house_name": fields.String(),
 })
data_in_json_serialize =  marshal(db_data, data_marshal_obj)}
print(type(data_in_json_serialize )) #  <class 'dict'>

定制序列化编组在烧瓶restpluse

你可以把你的对象输出为一个字典:

class User:
   def as_dict(self):
       return {c.name: getattr(self, c.name) for c in self.__table__.columns}

然后使用User.as_dict()序列化对象。

如将sqlalchemy行对象转换为python dict中所述