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

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'}}))

其他回答

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)

如果你正在使用Flask并且只想快速查询:

def get_cats():
    sql = text("select * from cat")
    sql_params = {}
    result = db.session.execute(sql, sql_params)
    row_list = result.fetchall()
    data = [dict(r) for r in row_list]

    response = jsonify({
        'data': [{
            'categorias': data
        }]
    })
    
    return response

(Sasha B的回答非常棒)

这特别地将datetime对象转换为字符串,在原始答案中将转换为None:

# Standard library imports
from datetime import datetime
import json

# 3rd party imports
from sqlalchemy.ext.declarative import DeclarativeMeta

class JsonEncoder(json.JSONEncoder):
    def default(self, obj):
        if isinstance(obj.__class__, DeclarativeMeta):
            dict = {}

            # Remove invalid fields and just get the column attributes
            columns = [x for x in dir(obj) if not x.startswith("_") and x != "metadata"]

            for column in columns:
                value = obj.__getattribute__(column)

                try:
                    json.dumps(value)
                    dict[column] = value
                except TypeError:
                    if isinstance(value, datetime):
                        dict[column] = value.__str__()
                    else:
                        dict[column] = None
            return dict

        return json.JSONEncoder.default(self, obj)

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。

def alc2json(row):
    return dict([(col, str(getattr(row,col))) for col in row.__table__.columns.keys()])

我想和她玩会儿代码高尔夫。

供参考:我使用automap_base,因为我们有一个根据业务需求单独设计的模式。我今天才开始使用SQLAlchemy,但是文档指出automap_base是declarative_base的扩展,这似乎是SQLAlchemy ORM中的典型范例,所以我相信这应该可以工作。

根据Tjorriemorrie的解决方案,它并没有跟随外键,而是简单地将列与值匹配,并通过str()-ing列值来处理Python类型。我们的值包括Python datetime。时间和小数。十进位类类型的结果,所以它完成了工作。

希望对路人有所帮助!