在Django Rest框架中,有没有一种方法可以聚合一个字段的总和
目前我已经知道如何在我的序列化程序中聚合一个列。py但是我的工资总额字段将进入序列化程序中的总工资字段,而总工资不在我的模型中。现在我的问题是如何在下面的API中执行类似操作:在Django Rest框架中,有没有一种方法可以聚合一个字段的总和,django,django-rest-framework,django-rest-viewsets,Django,Django Rest Framework,Django Rest Viewsets,目前我已经知道如何在我的序列化程序中聚合一个列。py但是我的工资总额字段将进入序列化程序中的总工资字段,而总工资不在我的模型中。现在我的问题是如何在下面的API中执行类似操作: "total_salary": 1422.05, { "id": "8c1810d9-b799-46a9-8506-3c18ef0067f8", "date": "2019-04-27", "virtual_assistant": "Joevie", "ti
"total_salary": 1422.05,
{
"id": "8c1810d9-b799-46a9-8506-3c18ef0067f8",
"date": "2019-04-27",
"virtual_assistant": "Joevie",
"time_in": "2019-04-27T22:20:13+08:00",
"time_out": "2019-04-28T05:20:13+08:00",
"hours": "7.00",
"client_name": "landmaster",
"rate": "90.00",
"salary": "630.00",
"status": "APPROVED-BY-THE-MANAGER",
"notes": ""
},
现有的是这样的
{
"id": "8c1810d9-b799-46a9-8506-3c18ef0067f8",
"total_salary": 1422.05,
"date": "2019-04-27",
"virtual_assistant": "Joevie",
"time_in": "2019-04-27T22:20:13+08:00",
"time_out": "2019-04-28T05:20:13+08:00",
"hours": "7.00",
"client_name": "landmaster",
"rate": "90.00",
"salary": "630.00",
"status": "APPROVED-BY-THE-MANAGER",
"notes": ""
},
目前我所做的一个解决方法是,我在ListView中获得截至目前的工资总额,并且每次用户都会搜索特定月份。计算将根据用户搜索的月份进行更改。代码如下
def get(self, request, *args, **kwargs):
search = request.GET.get('search')
user = request.user.staffs.full_name
current_month = datetime.date.today().month
current_year = datetime.date.today().year
payroll_list = VaPayroll.objects.all()
payroll_data = payroll_list.filter(Q(virtual_assistant=user),
Q(date__month=current_month),
Q(status='APPROVED-BY-THE-MANAGER'))
total_salary = VaPayroll.objects.filter(Q(virtual_assistant=user),
Q(date__month=current_month),
Q(status='APPROVED-BY-THE-MANAGER'),
Q(date__year=current_year)).aggregate(Sum('salary'))
if search:
payroll_data = payroll_list.filter(Q(virtual_assistant=user),
Q(status='APPROVED-BY-THE-MANAGER'),
Q(date__icontains=search))
total_salary = VaPayroll.objects.filter(Q(virtual_assistant=user),
Q(status='APPROVED-BY-THE-MANAGER'),
Q(date__month=search),
Q(date__year=current_year)).aggregate(Sum('salary'))
context = {
'total_salary': total_salary,
'payroll_data': payroll_data
}
return render(request, self.template_name, context)
这是我的serializers.py
class VaPayrollSerializer(serializers.ModelSerializer):
total_salary = serializers.SerializerMethodField()
class Meta:
model = VaPayroll
fields = '__all__'
def get_total_salary(self, obj):
user = self.context['request'].user.staffs.full_name
totalsalary = VaPayroll.objects.filter(Q(status='APPROVED-BY-THE-MANAGER'),
Q(virtual_assistant=user),
Q(date__month=datetime.date.today().month),
Q(date__year=datetime.date.today().year)).aggregate(total_salary=Sum('salary'))
return totalsalary['total_salary']
这是我的模型视图集aget\u queryset
def get_queryset(self):
current_month = datetime.date.today().month
current_year = datetime.date.today().year
queryset = VaPayroll.objects.filter(Q(virtual_assistant=self.request.user.staffs.full_name),
Q(date__month=current_month),
Q(date__year=current_year))
return queryset
在我的计划用例中,我希望在API中有一列包含所有薪资的总和,这样我就可以在前端单独调用该字段,而无需刷新页面来重新计算总薪资。还是这是正确的方法?或者我应该坚持只重新计算总工资的视图。你不能按你想要的方式获取json,因为它不是有效的json格式。我的建议是尝试以下方法:
"salary": {
"total": 1422.05,
"detail": {
"id": "8c1810d9-b799-46a9-8506-3c18ef0067f8",
"date": "2019-04-27",
"virtual_assistant": "Joevie",
"time_in": "2019-04-27T22:20:13+08:00",
"time_out": "2019-04-28T05:20:13+08:00",
"hours": "7.00",
"client_name": "landmaster",
"rate": "90.00",
"salary": "630.00",
"status": "APPROVED-BY-THE-MANAGER",
"notes": ""
}
}
为此,您必须使用to_representation()
方法指示序列化程序:
class VaPayrollSerializer(serializers.ModelSerializer):
class Meta:
model = VaPayroll
fields = '__all__'
def to_representation(self, instance):
original_representation = super().to_representation(instance)
representation = {
'total': self.get_total_salary(instance),
'detail': original_representation,
}
return representation
def get_total_salary(self, obj):
user = self.context['request'].user.staffs.full_name
totalsalary = VaPayroll.objects.filter(Q(status='APPROVED-BY-THE-MANAGER'),
Q(virtual_assistant=user),
Q(date__month=datetime.date.today().month),
Q(date__year=datetime.date.today().year)).aggregate(total_salary=Sum('salary'))
return totalsalary['total_salary']