删除 Mongo $ 登录 JSON

Question

我正在使用 MongoDB.

为 SPA (Angular) 构建 python 后端

这是我使用的：Python 3.4、MongoDB 3、Flask、flask-mongoengine 和 flask-restful

现在我从后端收到以下 JSON：

[
    {
        "_id": {
            "$oid": "55c737029380f82fbf52eec3"
        },
        "created_at": {
            "$date": 1439129906376
        },
        "desc": "Description.....",
        "title": "This is title"
    },
    etc...
]

还有我想收到类似的东西:

[
    {
        "_id": "55c737029380f82fbf52eec3",
        "created_at": 1439129906376,
        "desc": "Description.....",
        "title": "This is title"
    },
    etc...
]

我现在的代码：

from flask import json
from vinnie import app
from flask_restful import Resource, Api
from vinnie.models.movie import Movie

api = Api(app)

class Movies(Resource):
    def get(self):
        movies = json.loads(Movie.objects().all().to_json())
        return movies

api.add_resource(Movies, '/movies')

型号：

import datetime
from vinnie import db

class Movie(db.Document):
    created_at = db.DateTimeField(default=datetime.datetime.now, required=True)
    title = db.StringField(max_length=255, required=True)
    desc = db.StringField(required=True)

    def __unicode__(self):
        return self.title

格式化方便JSON前端的最佳方式是什么？

Answer 1

如果您有信心摆脱所有类似的情况，那么您当然可以编写符合该模式的代码。例如：

info = [
    {
        "_id": {
            "$oid": "55c737029380f82fbf52eec3"
        },
        "created_at": {
            "$date": 1439129906376
        },
        "desc": "Description.....",
        "title": "This is title"
    },
    #etc...
]

def fix_array(info):
    ''' Change out dict items in the following case:
           - dict value is another dict
           - the sub-dictionary only has one entry
           - the key in the subdictionary starts with '$'
        In this specific case, one level of indirection
        is removed, and the dict value is replaced with
        the sub-dict value.
    '''
    for item in info:
        for key, value in item.items():
            if not isinstance(value, dict) or len(value) != 1:
                continue
            (subkey, subvalue), = value.items()
            if not subkey.startswith('$'):
                continue
            item[key] = subvalue

fix_array(info)
print(info)

这将 return 这:

[{'title': 'This is title', 'created_at': 1439129906376, 'desc': 'Description.....', '_id': '55c737029380f82fbf52eec3'}]

显然，用 JSON 重新格式化是微不足道的。

Answer 2

我在我使用的 flask-restful 扩展程序中找到了解决我的问题的巧妙方法。

提供fields模块。

Flask-RESTful provides an easy way to control what data you actually render in your response. With the fields module, you can use whatever objects (ORM models/custom classes/etc.) you want in your resource. fields also lets you format and filter the response so you don’t have to worry about exposing internal data structures.

It’s also very clear when looking at your code what data will be rendered and how it will be formatted.

示例：

from flask_restful import Resource, fields, marshal_with

resource_fields = {
    'name': fields.String,
    'address': fields.String,
    'date_updated': fields.DateTime(dt_format='rfc822'),
}

class Todo(Resource):
    @marshal_with(resource_fields, envelope='resource')
    def get(self, **kwargs):
        return db_get_todo()  # Some function that queries the db

Flask-RESTful Output Fields Documentation

删除 Mongo $ 登录 JSON

Get rid of Mongo $ signs in JSON

json

mongodb

flask

python-3.4

flask-mongoengine