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

Flask-JsonTools包为您的模型提供了JsonSerializableBase基类的实现。

用法:

from sqlalchemy.ext.declarative import declarative_base
from flask.ext.jsontools import JsonSerializableBase

Base = declarative_base(cls=(JsonSerializableBase,))

class User(Base):
    #...

现在User模型可以神奇地序列化了。

如果你的框架不是Flask,你可以抓取代码

其他回答

我建议用棉花糖。它允许您创建序列化器来表示支持关系和嵌套对象的模型实例。

以下是他们文档中的一个删节的例子。以ORM模型为例,作者:

class Author(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    first = db.Column(db.String(80))
    last = db.Column(db.String(80))

该类的棉花糖模式是这样构造的:

class AuthorSchema(Schema):
    id = fields.Int(dump_only=True)
    first = fields.Str()
    last = fields.Str()
    formatted_name = fields.Method("format_name", dump_only=True)

    def format_name(self, author):
        return "{}, {}".format(author.last, author.first)

...并像这样使用:

author_schema = AuthorSchema()
author_schema.dump(Author.query.first())

...会产生这样的输出:

{
        "first": "Tim",
        "formatted_name": "Peters, Tim",
        "id": 1,
        "last": "Peters"
}

看看他们完整的Flask-SQLAlchemy示例。

一个名为marshmlow - SQLAlchemy的库专门集成了SQLAlchemy和marshmallow。在这个库中,上面描述的Author模型的模式如下所示:

class AuthorSchema(ModelSchema):
    class Meta:
        model = Author

该集成允许从SQLAlchemy Column类型推断字段类型。

marshmallow-sqlalchemy这里。

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)

2023年末

我的实现

def obj_to_dict(obj, remove=['_sa_instance_state'], debug=False):
    result = {}

    if type(obj).__name__ == "Row":
        return dict(obj)

    obj = obj.__dict__
    for key in obj:
        if key in remove:
            continue

        result[key] = obj[key]

    if debug:
        print(result)

    return result

扁平化实现

你可以使用这样的代码:

from sqlalchemy.ext.declarative import DeclarativeMeta

class AlchemyEncoder(json.JSONEncoder):

    def default(self, obj):
        if isinstance(obj.__class__, DeclarativeMeta):
            # an SQLAlchemy class
            fields = {}
            for field in [x for x in dir(obj) if not x.startswith('_') and x != 'metadata']:
                data = obj.__getattribute__(field)
                try:
                    json.dumps(data) # this will fail on non-encodable values, like other classes
                    fields[field] = data
                except TypeError:
                    fields[field] = None
            # a json-encodable dict
            return fields

        return json.JSONEncoder.default(self, obj)

然后转换为JSON使用:

c = YourAlchemyClass()
print json.dumps(c, cls=AlchemyEncoder)

它将忽略不可编码的字段(将它们设置为“None”)。

它不会自动展开关系(因为这可能导致自引用,并永远循环)。

递归的非循环实现

然而,如果你宁愿永远循环,你可以使用:

from sqlalchemy.ext.declarative import DeclarativeMeta

def new_alchemy_encoder():
    _visited_objs = []

    class AlchemyEncoder(json.JSONEncoder):
        def default(self, obj):
            if isinstance(obj.__class__, DeclarativeMeta):
                # don't re-visit self
                if obj in _visited_objs:
                    return None
                _visited_objs.append(obj)

                # an SQLAlchemy class
                fields = {}
                for field in [x for x in dir(obj) if not x.startswith('_') and x != 'metadata']:
                    fields[field] = obj.__getattribute__(field)
                # a json-encodable dict
                return fields

            return json.JSONEncoder.default(self, obj)

    return AlchemyEncoder

然后对对象进行编码,使用:

print json.dumps(e, cls=new_alchemy_encoder(), check_circular=False)

这将编码所有的子代、子代、子代……基本上可以编码你的整个数据库。当它到达之前编码过的东西时,它会将其编码为“None”。

递归的、可能是循环的、有选择的实现

另一种选择,可能更好,是能够指定你想要展开的字段:

def new_alchemy_encoder(revisit_self = False, fields_to_expand = []):
    _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']:
                    val = obj.__getattribute__(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] = None
                            continue

                    fields[field] = val
                # a json-encodable dict
                return fields

            return json.JSONEncoder.default(self, obj)

    return AlchemyEncoder

你现在可以调用它:

print json.dumps(e, cls=new_alchemy_encoder(False, ['parents']), check_circular=False)

例如,仅展开名为“parents”的SQLAlchemy字段。

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'}}))