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/)
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
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'}}))
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