Question

我从Order对象的实例开始，并尝试将它们转换为JSON格式，这将用于更新表。我是特定于Django的代码构造的新手。

我选择的方法是：

将数据从数据库加载为多个Order实例
将Order个实例转换为中间Table对象（格式已赋予）
将Table对象序列化为JSON

到目前为止我走得很远，但除了在Python shell中运行之外，我无法运行整个应用程序。我已经有了模型和序列化器。

有人可以提供他的帮助，如何一次标记所有Order个实例，并在一次API调用中将它们转换为Table以及此简单示例中缺少的任何其他内容吗？

models.py：

# Order corresponds to Line in the Table
class Order(models.Model):
    doc = models.CharField(max_length=200, blank=True, null=True)
    order = models.CharField(max_length=200, blank=True, null=True)
    nothing = models.CharField(max_length=200, blank=True, null=True)

    def __str__(self):
        return self.order

class Table(models.Model):
    pass

class Column(models.Model):
    data = models.CharField(max_length=200, blank=True, null=True)
    table = models.ForeignKey(Table)

    def __str__(self):
        return self.data

class Line(models.Model):
    doc = models.CharField(max_length=200, blank=True, null=True)
    order = models.CharField(max_length=200, blank=True, null=True)
    nothing = models.CharField(max_length=200, blank=True, null=True)
    table = models.ForeignKey(Table)

    def __str__(self):
        return self.order

serializers.py：

class TableSerializer(serializers.ModelSerializer):

    columns = ColumnSerializer(many=True)
    lines = LineSerializer(many=True)

    class Meta:
        model = Table
        fields = [
            'columns',
            'lines'
        ]

class ColumnSerializer(serializers.ModelSerializer):

    class Meta:
        model = Column
        fields = [
            'data'
        ]

class LineSerializer(serializers.ModelSerializer):

    class Meta:
        model = Line
        fields = [
            'doc',
            'order',
            'nothing'
        ]

对于此关系数据：

doc         order       nothing
564251422   564210      5648
546546545   98745       4668

JSON输出应为：

{
    "columns": [
        {
            "data": "doc"
        },
        {
            "data": "order"
        },
        {
            "data": "nothing"
        }
    ],
    "lines": [
        {
            "doc": "564251422",
            "nothing": 0.0,
            "order": "56421"
        },
        {
            "doc": "546546545",
            "nothing": 0.0,
            "order": "98745"
        }
    ]
}

Shell命令：

import polls.models
polls.models.Order.objects.all() # outputs all orders
table = polls.models.Table()
polls.models.Column(data="doc",table=table)
polls.models.Column(data="order",table=table)
polls.models.Column(data="nothing",table=table)
polls.models.Line(doc="564251422",order="56421",nothing="0.0",table=table)
polls.models.Line(doc="546546545",order="98745",nothing="0.0",table=table)

views.py（更新）：

bunch = OrderSerializer(Order.objects.all(), many=True)                                                                              
headers = bunch.data[0].keys()                                                                                                                   
headers_prepared = map (lambda x: {'data': x} , headers)
ordered_all = ( ('columns', headers_prepared), ('lines', bunch.data) )
data = collections.OrderedDict(ordered_all)
data_json = JSONRenderer().render(data)
return JsonResponse(data_json, safe=False)

Answer 1

我认为这个问题可能与另一个问题有关（在评论中提到），但对于这个特殊情况......

当您使用DjangoRestFramework（正确的方式）时，要获得Json格式的所有数据，您可以：

serialisers.py中的

：

from rest_framework import serializers

from polls.models import Order

class OrderSerializer(serializers.ModelSerializer):
    class Meta:
        model = Order
        fields = ('doc', 'order', 'nothing')

# note we do not use other serializers

下一个 shell （将来的视图）：

from rest_framework.renderers import JSONRenderer
from polls.models import Order
from polls.serializers import OrderSerializer

bunch = OrderSerializer(Order.objects.all(), many=True)

#this will output "lines" part of desired output
JSONRenderer().render(bunch.data)

#next to get columns, or i say headers
headers = bunch.data[0].keys() 
# headers = ['doc','order','nothing']
# !note will fail if bunch.data is empty

headers_prepared = map (lambda x: {'data': x} , headers)
# headers_prepared = [{'data': 'doc'}, {'data': 'order'}, {'data': 'nothing'}]

import collections # need to use OrderedDict to store in our sequence
ordered_all = ( ('columns', headers_prepared), ('lines', bunch.data) )

#finally desired output 
JSONRenderer().render( collections.OrderedDict(ordered_all)   )

#all code and output tested on my dummy data

urls.py中的

更新：

urlpatterns = [
    ...
    url(r'my/', my),
    ...
]

添加views.py：

#as you sileny JSONResponse (from RestFramework docs) - need to show it 
from django.http import HttpResponse

class JSONResponse(HttpResponse):
"""
An HttpResponse that renders its content into JSON.
"""
    def __init__(self, data, **kwargs):
        content = JSONRenderer().render(data)
        kwargs['content_type'] = 'application/json'
        super(JSONResponse, self).__init__(content, **kwargs)

#now our view

import collections 

def my(request):

    bunch = OrderSerializer(Order.objects.all(), many=True)

    # to get headers independent of presence of orders
    empty = OrderSerializer()
    headers = empty.data.keys()

    headers_prepared = map (lambda x: {'data': x} , headers)
    ordered_all = ( ('columns', headers_prepared), ('lines', bunch.data) )

    out = collections.OrderedDict(ordered_all)
    #finally desired output 
    return  JSONResponse( out  )

将数据导入特定模型

1 个答案: