打破 django json 响应

Question

我对代码有以下响应：

{
    "id": 716,
    "name": "XYZ",
    "start_date": "2019-12-24",
    "end_date": "2020-01-31",        
    "ads": [
        {
            "id": 20228,
            "no_of_times_per_hr": 1,
        },
        {
            "id": 20227,
            "no_of_times_per_hr": 2
        },
        {
            "id": 20229,
            "no_of_times_per_hr": 7                
        }
    ]
},

我使用序列化程序编写了以下代码：

class AdsDetailOnScheduler(serializers.ModelSerializer):
    ads = serializers.SerializerMethodField()
    class Meta:
      model = AdCampaign
      fields = ('id','name','start_date','end_date', 'ads')

    def get_ads(self, obj):
      adlabels = AdDayLabelMap.objects.filter(ad_campaign__id  = obj['id']).values_list('ad_label', flat = True).distinct()
      ad_id = AdSlot.objects.filter(ad_label__id__in = adlabels).values_list('ads', flat = True)
      if Ads.objects.filter(id__in = ad_id).exists():
        a = []
        for ad in Ads.objects.filter(id__in = ad_id):
            a.append(AdsDetailSongScheduler(ad).data)
        return a
      else:
        return None  

class AdsDetailSongScheduler(serializers.ModelSerializer):  
    no_of_times_per_hr = serializers.SerializerMethodField()
    after_n_songs = serializers.SerializerMethodField()
    specific_time = serializers.SerializerMethodField()
    class Meta:
      model = Ads
      fields = ('id','no_of_times_per_hr')

    def get_no_of_times_per_hr(self, obj):          
      if obj.no_of_times_per_hr:
         return obj.no_of_times_per_hr
      else:
         return None

但是，我希望得到以下答复：

{
    "id": 716,
    "name": "XYZ",
    "start_date": "2019-12-24",
    "end_date": "2020-01-31",        
    "ads": [
        {
            "id": 20228,
            "no_of_times_per_hr": 1,
        },
    ]
},



{
    "id": 716,
    "name": "XYZ",
    "start_date": "2019-12-24",
    "end_date": "2020-01-31",        
    "ads": [
        {
            "id": 20227,
            "no_of_times_per_hr": 2
        },
    ]
},



{
    "id": 716,
    "name": "XYZ",
    "start_date": "2019-12-24",
    "end_date": "2020-01-31",        
    "ads": [
        {
            "id": 20229,
            "no_of_times_per_hr": 7                
        }
    ]
},

我想解释一下，广告是以列表形式出现在单个响应中的。我想用相同的名称、start_date、end_date 值打破广告。有可能吗？

更新 1：

之后，我得到的回复之一，

[
    {
        "end_date": "2020-01-31",
        "id": 76,
        "ads": [
            {
                "id": 208,                    
                "no_of_times_per_hr": 1
            }
        ],
        "name": "XYZ",
        "start_date": "2019-12-24"
    },
    {
        "end_date": "2020-01-31",
        "id": 76,
        "ads": [
            {
                "id": 207,                
                "no_of_times_per_hr": 2
            }
        ],
        "name": "XYZ",            
        "start_date": "2019-12-24"
    },
    {
        "end_date": "2020-01-31",
        "id": 76,
        "ads": [
            {
                "id": 209,                    
                "no_of_times_per_hr": 7
            }
        ],
        "name": "XYZ",
        "start_date": "2019-12-24"
    }
],

上面的响应算作一个 JSON 响应而不是三个，因为响应是一个列表，而不是三个不同的列表。那么，是否有人可以帮忙？

Answer 1

在序列化器级别，您可以通过覆盖AdsDetailOnScheduler 类中的to_representation 方法来实现（下面的示例）。

但是因为消费者而改变响应并不是一个好习惯。 API 使用者应根据需要处理 API 响应。如果您可以访问消费者代码，我建议您将其更改为破坏对象关系。

def to_representation(self, instance):
    response_dict = super().to_representation(instance)
    ads = response_dict.pop('ads', [])
    output_list = []
    for ad in ads:
        data = {}
        data.update(response_dict)
        data['ads'] = [ad]
        output_list.append(data)
    return output_list