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
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
其他回答
(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)
虽然使用一些原始sql和未定义的对象，使用cursor.description似乎得到了我正在寻找的东西:
with connection.cursor() as cur:
print(query)
cur.execute(query)
for item in cur.fetchall():
row = {column.name: item[i] for i, column in enumerate(cur.description)}
print(row)
step1:
class CNAME:
...
def as_dict(self):
return {item.name: getattr(self, item.name) for item in self.__table__.columns}
step2:
list = []
for data in session.query(CNAME).all():
list.append(data.as_dict())
step3:
return jsonify(list)
内置序列化器因utf-8而阻塞，无法解码某些输入的无效开始字节。相反，我的答案是:
def row_to_dict(row):
temp = row.__dict__
temp.pop('_sa_instance_state', None)
return temp
def rows_to_list(rows):
ret_rows = []
for row in rows:
ret_rows.append(row_to_dict(row))
return ret_rows
@website_blueprint.route('/api/v1/some/endpoint', methods=['GET'])
def some_api():
'''
/some_endpoint
'''
rows = rows_to_list(SomeModel.query.all())
response = app.response_class(
response=jsonplus.dumps(rows),
status=200,
mimetype='application/json'
)
return response
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
