python - 对 Django 中的查询集中的一个字段求​​和

Question

模型.py

class QaCommission(models.Model):
    user = models.ForeignKey(PlUser, on_delete=models.CASCADE, blank=True, null=True, related_name='user_commission')
    ref = models.ForeignKey(PlUser, on_delete=models.CASCADE, blank=True, null=True, related_name='ref_commission')
    price = models.FloatField(blank=True, null=True)
    pct = models.FloatField(blank=True, null=True)
    commission = models.FloatField(blank=True, null=True)
    status = models.IntegerField(blank=True, null=True, default=0)

序列化器.py

class QaCommissionSerializer(serializers.ModelSerializer):
    class Meta:
        model = QaCommission
        fields = '__all__'

View .py

class QaCommissionList(viewsets.ModelViewSet):
    queryset = QaCommission.objects.all()
    serializer_class = QaCommissionSerializer

如果我们在此 View 中过​​滤 ref=60，结果显示如下:

{
"count": 18,
"next": "http://127.0.0.1:8008/api/qacommission/?ref=60&page=2",
"previous": null,
"results": [
    {
        "id": 1,
        "price": 20.0,
        "pct": 0.1,
        "commission": 2.0,
        "status": 1,
        "user": 7,
        "ref": 60
    },
    {
        "id": 2,
        "price": 10.0,
        "pct": 0.1,
        "commission": 1.0,
        "status": 1,
        "user": 7,
        "ref": 60
    },
    ......
    ......
    ......
    {
        "id": 10,
        "price": 15.0,
        "pct": 0.1,
        "commission": 1.5,
        "status": 1,
        "user": 7,
        "ref": 60
    }
]
}

我想对结果中所有的“commission”字段求和，并将总和附加到原始查询集(可能在“count”:18旁边)，如上所示，有18个佣金需要计算。

我该如何实现这个？需要您的帮助，谢谢!

Answer 1

尝试将 ModelViewset 的 list() 方法重写为，

class QaCommissionList(viewsets.ModelViewSet):
    queryset = QaCommission.objects.all()
    serializer_class = QaCommissionSerializer

    <b>def list(self, request, *args, **kwargs):
        response = super().list(request, *args, **kwargs)
        response.data['sum'] = sum([data.get('commission', 0) for data in response.data['results']])
        return response</b>

此答案总结了特定页面的佣金并显示出来

更新

<b>from django.db.models import Sum</b>


class QaCommissionList(viewsets.ModelViewSet):
    queryset = QaCommission.objects.all()
    serializer_class = QaCommissionSerializer

    <b>def list(self, request, *args, **kwargs):
        response = super().list(request, *args, **kwargs)
        if 'ref' in request.GET and request.GET['ref']:
            response.data['sum'] = QaCommission.objects.filter(ref=int(request.GET['ref'])
                                                               ).aggregate(sum=Sum('commission'))['sum']
        return response</b>

上述答案将返回过滤器的commission列的总和(与分页无关)

谢谢@bruno提到这样一个有效的观点