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

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

其他回答

下面是一个解决方案,它允许您选择希望在输出中包含的关系。 注意:这是一个完整的重写,将dict/str作为一个参数,而不是一个列表。修复了一些东西..

def deep_dict(self, relations={}):
    """Output a dict of an SA object recursing as deep as you want.

    Takes one argument, relations which is a dictionary of relations we'd
    like to pull out. The relations dict items can be a single relation
    name or deeper relation names connected by sub dicts

    Example:
        Say we have a Person object with a family relationship
            person.deep_dict(relations={'family':None})
        Say the family object has homes as a relation then we can do
            person.deep_dict(relations={'family':{'homes':None}})
            OR
            person.deep_dict(relations={'family':'homes'})
        Say homes has a relation like rooms you can do
            person.deep_dict(relations={'family':{'homes':'rooms'}})
            and so on...
    """
    mydict =  dict((c, str(a)) for c, a in
                    self.__dict__.items() if c != '_sa_instance_state')
    if not relations:
        # just return ourselves
        return mydict

    # otherwise we need to go deeper
    if not isinstance(relations, dict) and not isinstance(relations, str):
        raise Exception("relations should be a dict, it is of type {}".format(type(relations)))

    # got here so check and handle if we were passed a dict
    if isinstance(relations, dict):
        # we were passed deeper info
        for left, right in relations.items():
            myrel = getattr(self, left)
            if isinstance(myrel, list):
                mydict[left] = [rel.deep_dict(relations=right) for rel in myrel]
            else:
                mydict[left] = myrel.deep_dict(relations=right)
    # if we get here check and handle if we were passed a string
    elif isinstance(relations, str):
        # passed a single item
        myrel = getattr(self, relations)
        left = relations
        if isinstance(myrel, list):
            mydict[left] = [rel.deep_dict(relations=None)
                                 for rel in myrel]
        else:
            mydict[left] = myrel.deep_dict(relations=None)

    return mydict

举个关于person/family/homes/rooms的例子…把它转换成json,你只需要

json.dumps(person.deep_dict(relations={'family':{'homes':'rooms'}}))

这是一个JSONEncoder版本,它保留了模型列的顺序,只保留递归定义的列和关系字段。它还格式化了大多数不可序列化的JSON类型:

import json
from datetime import datetime
from decimal import Decimal

import arrow
from sqlalchemy.ext.declarative import DeclarativeMeta

class SQLAlchemyJSONEncoder(json.JSONEncoder):
    """
    SQLAlchemy ORM JSON Encoder
    If you have a "backref" relationship defined in your SQLAlchemy model,
    this encoder raises a ValueError to stop an infinite loop.
    """

    def default(self, obj):
        if isinstance(obj, datetime):
            return arrow.get(obj).isoformat()
        elif isinstance(obj, Decimal):
            return float(obj)
        elif isinstance(obj, set):
            return sorted(obj)
        elif isinstance(obj.__class__, DeclarativeMeta):
            for attribute, relationship in obj.__mapper__.relationships.items():
                if isinstance(relationship.__getattribute__("backref"), tuple):
                    raise ValueError(
                        f'{obj.__class__} object has a "backref" relationship '
                        "that would cause an infinite loop!"
                    )
            dictionary = {}
            column_names = [column.name for column in obj.__table__.columns]
            for key in column_names:
                value = obj.__getattribute__(key)
                if isinstance(value, datetime):
                    value = arrow.get(value).isoformat()
                elif isinstance(value, Decimal):
                    value = float(value)
                elif isinstance(value, set):
                    value = sorted(value)
                dictionary[key] = value
            for key in [
                attribute
                for attribute in dir(obj)
                if not attribute.startswith("_")
                and attribute != "metadata"
                and attribute not in column_names
            ]:
                value = obj.__getattribute__(key)
                dictionary[key] = value
            return dictionary

        return super().default(obj)

在SQLAlchemy中使用内置序列化器:

from sqlalchemy.ext.serializer import loads, dumps
obj = MyAlchemyObject()
# serialize object
serialized_obj = dumps(obj)

# deserialize object
obj = loads(serialized_obj)

如果在会话之间传输对象,请记住使用session.expunge(obj)将对象从当前会话中分离出来。 要再次附加它,只需执行session.add(obj)。

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

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