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/)
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)
我建议用棉花糖。它允许您创建序列化器来表示支持关系和嵌套对象的模型实例。
以下是他们文档中的一个删节的例子。以ORM模型为例,作者:
class Author(db.Model):
id = db.Column(db.Integer, primary_key=True)
first = db.Column(db.String(80))
last = db.Column(db.String(80))
该类的棉花糖模式是这样构造的:
class AuthorSchema(Schema):
id = fields.Int(dump_only=True)
first = fields.Str()
last = fields.Str()
formatted_name = fields.Method("format_name", dump_only=True)
def format_name(self, author):
return "{}, {}".format(author.last, author.first)
...并像这样使用:
author_schema = AuthorSchema()
author_schema.dump(Author.query.first())
...会产生这样的输出:
{
"first": "Tim",
"formatted_name": "Peters, Tim",
"id": 1,
"last": "Peters"
}
看看他们完整的Flask-SQLAlchemy示例。
一个名为marshmlow - SQLAlchemy的库专门集成了SQLAlchemy和marshmallow。在这个库中,上面描述的Author模型的模式如下所示:
class AuthorSchema(ModelSchema):
class Meta:
model = Author
该集成允许从SQLAlchemy Column类型推断字段类型。
marshmallow-sqlalchemy这里。
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'}}))
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