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
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'))
经过一番尝试，我想出了自己的解决方案
def to_dict(self):
keys = self.__mapper__.attrs.keys()
attrs = vars(self)
return { k : attrs[k] for k in keys}
Python 3.7+和Flask 1.1+可以使用内置的数据类包
from dataclasses import dataclass
from datetime import datetime
from flask import Flask, jsonify
from flask_sqlalchemy import SQLAlchemy
app = Flask(__name__)
db = SQLAlchemy(app)
@dataclass
class User(db.Model):
id: int
email: str
id = db.Column(db.Integer, primary_key=True, auto_increment=True)
email = db.Column(db.String(200), unique=True)
@app.route('/users/')
def users():
users = User.query.all()
return jsonify(users)
if __name__ == "__main__":
users = User(email="user1@gmail.com"), User(email="user2@gmail.com")
db.create_all()
db.session.add_all(users)
db.session.commit()
app.run()
/users/路由现在将返回一个用户列表。
[
{"email": "user1@gmail.com", "id": 1},
{"email": "user2@gmail.com", "id": 2}
]
自动序列化相关模型
@dataclass
class Account(db.Model):
id: int
users: User
id = db.Column(db.Integer)
users = db.relationship(User) # User model would need a db.ForeignKey field
jsonify(account)的响应是这样的。
{
"id":1,
"users":[
{
"email":"user1@gmail.com",
"id":1
},
{
"email":"user2@gmail.com",
"id":2
}
]
}
覆盖默认的JSON编码器
from flask.json import JSONEncoder
class CustomJSONEncoder(JSONEncoder):
"Add support for serializing timedeltas"
def default(o):
if type(o) == datetime.timedelta:
return str(o)
if type(o) == datetime.datetime:
return o.isoformat()
return super().default(o)
app.json_encoder = CustomJSONEncoder
安装simplejson by PIP安装simplejson并创建一个类
class Serialise(object):
def _asdict(self):
"""
Serialization logic for converting entities using flask's jsonify
:return: An ordered dictionary
:rtype: :class:`collections.OrderedDict`
"""
result = OrderedDict()
# Get the columns
for key in self.__mapper__.c.keys():
if isinstance(getattr(self, key), datetime):
result["x"] = getattr(self, key).timestamp() * 1000
result["timestamp"] = result["x"]
else:
result[key] = getattr(self, key)
return result
并将这个类继承到每个orm类，这样这个_asdict函数就会注册到每个orm类，然后。 并在任何地方使用jsonify
(Sasha B的回答非常棒)
这特别地将datetime对象转换为字符串，在原始答案中将转换为None:
# Standard library imports
from datetime import datetime
import json
# 3rd party imports
from sqlalchemy.ext.declarative import DeclarativeMeta
class JsonEncoder(json.JSONEncoder):
def default(self, obj):
if isinstance(obj.__class__, DeclarativeMeta):
dict = {}
# Remove invalid fields and just get the column attributes
columns = [x for x in dir(obj) if not x.startswith("_") and x != "metadata"]
for column in columns:
value = obj.__getattribute__(column)
try:
json.dumps(value)
dict[column] = value
except TypeError:
if isinstance(value, datetime):
dict[column] = value.__str__()
else:
dict[column] = None
return dict
return json.JSONEncoder.default(self, obj)
