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/)
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'))
我知道这是一个相当老的帖子。我采取了@SashaB给出的解决方案,并根据我的需要进行了修改。
我添加了以下内容:
字段忽略列表:序列化时要忽略的字段列表
字段替换列表:包含在序列化时要被值替换的字段名的字典。
删除方法和BaseQuery被序列化
我的代码如下:
def alchemy_json_encoder(revisit_self = False, fields_to_expand = [], fields_to_ignore = [], fields_to_replace = {}):
"""
Serialize SQLAlchemy result into JSon
:param revisit_self: True / False
:param fields_to_expand: Fields which are to be expanded for including their children and all
:param fields_to_ignore: Fields to be ignored while encoding
:param fields_to_replace: Field keys to be replaced by values assigned in dictionary
:return: Json serialized SQLAlchemy object
"""
_visited_objs = []
class AlchemyEncoder(json.JSONEncoder):
def default(self, obj):
if isinstance(obj.__class__, DeclarativeMeta):
# don't re-visit self
if revisit_self:
if obj in _visited_objs:
return None
_visited_objs.append(obj)
# go through each field in this SQLalchemy class
fields = {}
for field in [x for x in dir(obj) if not x.startswith('_') and x != 'metadata' and x not in fields_to_ignore]:
val = obj.__getattribute__(field)
# is this field method defination, or an SQLalchemy object
if not hasattr(val, "__call__") and not isinstance(val, BaseQuery):
field_name = fields_to_replace[field] if field in fields_to_replace else field
# is this field another SQLalchemy object, or a list of SQLalchemy objects?
if isinstance(val.__class__, DeclarativeMeta) or \
(isinstance(val, list) and len(val) > 0 and isinstance(val[0].__class__, DeclarativeMeta)):
# unless we're expanding this field, stop here
if field not in fields_to_expand:
# not expanding this field: set it to None and continue
fields[field_name] = None
continue
fields[field_name] = val
# a json-encodable dict
return fields
return json.JSONEncoder.default(self, obj)
return AlchemyEncoder
希望它能帮助到一些人!
更详细的解释。
在你的模型中,添加:
def as_dict(self):
return {c.name: str(getattr(self, c.name)) for c in self.__table__.columns}
str()是针对python3的,所以如果使用python2则使用unicode()。它应该有助于反序列化日期。如果不处理这些,你可以删除它。
现在可以像这样查询数据库
some_result = User.query.filter_by(id=current_user.id).first().as_dict()
需要First()来避免奇怪的错误。As_dict()现在将反序列化结果。反序列化之后,就可以将其转换为json了
jsonify(some_result)
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