[英]Sum up a field in the queryset in Django
models.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)
serializers.py
class QaCommissionSerializer(serializers.ModelSerializer):
class Meta:
model = QaCommission
fields = '__all__'
views.py
class QaCommissionList(viewsets.ModelViewSet):
queryset = QaCommission.objects.all()
serializer_class = QaCommissionSerializer
如果我们在此视图中过滤ref = 60,则结果显示如下:
{
"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
}
]
}
我想对结果中的所有“佣金”字段进行汇总,然后将总和附加到原始查询集(也许在“计数”旁边:18),如上所示,有18个佣金需要计数。
我该如何实施? 需要您的帮助,谢谢!
尝试覆盖ModelViewset
list()
方法,
class QaCommissionList(viewsets.ModelViewSet):
queryset = QaCommission.objects.all()
serializer_class = QaCommissionSerializer
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
这样的回答和行动的commision
从特定的页面 ,并显示它
UPDATE
from django.db.models import Sum
class QaCommissionList(viewsets.ModelViewSet):
queryset = QaCommission.objects.all()
serializer_class = QaCommissionSerializer
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
上面的答案将返回带有过滤器的“ commission
总额”列(与分页无关)
感谢@bruno提到这样的观点
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