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

AlchemyEncoder是很棒的,但有时会失败的十进制值。这是一个改进的编码器,解决十进制问题-

class AlchemyEncoder(json.JSONEncoder):
# To serialize SQLalchemy objects 
def default(self, obj):
    if isinstance(obj.__class__, DeclarativeMeta):
        model_fields = {}
        for field in [x for x in dir(obj) if not x.startswith('_') and x != 'metadata']:
            data = obj.__getattribute__(field)
            print data
            try:
                json.dumps(data)  # this will fail on non-encodable values, like other classes
                model_fields[field] = data
            except TypeError:
                model_fields[field] = None
        return model_fields
    if isinstance(obj, Decimal):
        return float(obj)
    return json.JSONEncoder.default(self, obj)

其他回答

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

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

下面的代码将sqlalchemy结果序列化为json。

import json
from collections import OrderedDict


def asdict(self):
    result = OrderedDict()
    for key in self.__mapper__.c.keys():
        if getattr(self, key) is not None:
            result[key] = str(getattr(self, key))
        else:
            result[key] = getattr(self, key)
    return result


def to_array(all_vendors):
    v = [ ven.asdict() for ven in all_vendors ]
    return json.dumps(v) 

叫有趣,

def all_products():
    all_products = Products.query.all()
    return to_array(all_products)

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)

我对使用(太多?)字典的看法:

def serialize(_query):
    #d = dictionary written to per row
    #D = dictionary d is written to each time, then reset
    #Master = dictionary of dictionaries; the id Key (int, unique from database) 
    from D is used as the Key for the dictionary D entry in Master
    Master = {}
    D = {}
    x = 0
    for u in _query:
        d = u.__dict__
        D = {}
        for n in d.keys():
           if n != '_sa_instance_state':
                    D[n] = d[n]
        x = d['id']
        Master[x] = D
    return Master

使用flask(包括jsonify)和flask_sqlalchemy将输出打印为JSON。

使用jsonify(serialize())调用该函数。

与我迄今为止尝试过的所有SQLAlchemy查询一起工作(运行SQLite3)