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/)
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'}}))
(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)
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'))