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

你可以像这样将RowProxy转换为dict:

 d = dict(row.items())

然后将其序列化为JSON(必须为datetime值等指定编码器) 如果您只想要一条记录(而不是相关记录的完整层次结构),这并不难。

json.dumps([(dict(row.items())) for row in rs])

其他回答

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'}}))

出于安全考虑,您不应该返回模型的所有字段。我更喜欢有选择性地选择他们。

Flask的json编码现在支持UUID, datetime和relationships(并为flask_sqlalchemy db添加了query和query_class。模型类)。编码器我更新如下:

app / json_encoder.py

    from sqlalchemy.ext.declarative import DeclarativeMeta
    from flask import json


    class AlchemyEncoder(json.JSONEncoder):
        def default(self, o):
            if isinstance(o.__class__, DeclarativeMeta):
                data = {}
                fields = o.__json__() if hasattr(o, '__json__') else dir(o)
                for field in [f for f in fields if not f.startswith('_') and f not in ['metadata', 'query', 'query_class']]:
                    value = o.__getattribute__(field)
                    try:
                        json.dumps(value)
                        data[field] = value
                    except TypeError:
                        data[field] = None
                return data
            return json.JSONEncoder.default(self, o)

app / __init__ . py

# json encoding
from app.json_encoder import AlchemyEncoder
app.json_encoder = AlchemyEncoder

有了这个,我可以选择添加一个__json__属性,返回我希望编码的字段列表:

app / models.py

class Queue(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    song_id = db.Column(db.Integer, db.ForeignKey('song.id'), unique=True, nullable=False)
    song = db.relationship('Song', lazy='joined')
    type = db.Column(db.String(20), server_default=u'audio/mpeg')
    src = db.Column(db.String(255), nullable=False)
    created_at = db.Column(db.DateTime, server_default=db.func.now())
    updated_at = db.Column(db.DateTime, server_default=db.func.now(), onupdate=db.func.now())

    def __init__(self, song):
        self.song = song
        self.src = song.full_path

    def __json__(self):
        return ['song', 'src', 'type', 'created_at']

我添加@jsonapi到我的视图,返回结果列表,然后我的输出如下:

[

{

    "created_at": "Thu, 23 Jul 2015 11:36:53 GMT",
    "song": 

        {
            "full_path": "/static/music/Audioslave/Audioslave [2002]/1 Cochise.mp3",
            "id": 2,
            "path_name": "Audioslave/Audioslave [2002]/1 Cochise.mp3"
        },
    "src": "/static/music/Audioslave/Audioslave [2002]/1 Cochise.mp3",
    "type": "audio/mpeg"
}

]

这并不是那么简单。我写了一些代码来做这件事。我还在开发中,它使用了MochiKit框架。它基本上使用代理和注册的JSON转换器在Python和Javascript之间转换复合对象。

数据库对象的浏览器端是db.js 它需要proxy.js中的基本Python代理源代码。

在Python方面,有基本代理模块。 最后是webserver.py中的SqlAlchemy对象编码器。 它还依赖于models.py文件中的元数据提取器。

你可以把你的对象输出为一个字典:

class User:
   def as_dict(self):
       return {c.name: getattr(self, c.name) for c in self.__table__.columns}

然后使用User.as_dict()序列化对象。

如将sqlalchemy行对象转换为python dict中所述

class SqlToDict:
    def __init__(self, data) -> None:
        self.data = data

    def to_timestamp(self, date):
        if isinstance(date, datetime):
            return int(datetime.timestamp(date))
        else:
            return date

    def to_dict(self) -> List:
        arr = []
        for i in self.data:
            keys = [*i.keys()]
            values = [*i]
            values = [self.to_timestamp(d) for d in values]
            arr.append(dict(zip(keys, values)))
        return arr

例如:

SqlToDict(data).to_dict()