在Django表单上,我想知道如何响应表单字段数据&x200B;JSON格式

在Django表单上,我想知道如何响应表单字段数据&x200B;JSON格式,django,forms,Django,Forms,我成功地使用了这个方法,但是,我希望RESTE代码更简单 有这样的方法吗 代码如下: views.py class PersonnelInfoDetailView(DetailView): def get(self, request, pk): employee = Employee.objects.get(pk=pk) response = {'employeeType' : employee.employeeType_id, 'driverIndex': empl

我成功地使用了这个方法,但是,我希望RESTE代码更简单

有这样的方法吗

代码如下:

views.py

class PersonnelInfoDetailView(DetailView):
      def get(self, request, pk):
    employee = Employee.objects.get(pk=pk)
    response = {'employeeType' : employee.employeeType_id, 'driverIndex': employee.driverIndex, 'routeId': employee.routeId_id, 'department': employee.department, 
    'employeeNumber': employee.employeeNumber, 'employeeName': employee.employeeName, 'employeeBirth' : employee.employeeBirth, 'employeeAddress' : employee.employeeAddress,
    'employeeTel' : employee.employeeTel, 'employeeEmergencyTel' : employee.employeeEmergencyTel, 'employeeSalary' : employee.employeeSalary, 'employeeFixedRest1' : employee.employeeFixedRest1,
    'employeeFixedRest2' : employee.employeeFixedRest2 , 'tableHoBong' : employee.tableHoBong_id, 'jobTitleType' : employee.jobTitleType, 'employeeStatus' : employee.employeeStatus }
    return HttpResponse(json.dumps(response), content_type='application/json')
forms.py

class EmployeeForm(forms.ModelForm):

class Meta:
    model = Employee
    fields = ('employeeType','driverIndex', 'routeId', 'department', 'employeeNumber', 'employeeName', 'employeeBirth', 'employeeAddress', 'employeeTel', 'employeeEmergencyTel', 'employeeSalary', 'employeeFixedRest1', 'employeeFixedRest2','tableHoBong','jobTitleType', 'employeeStatus')
def __init__(self, *args, **kwargs):
    super(EmployeeForm, self).__init__(*args, **kwargs)
    for visible in self.visible_fields():
        visible.field.widget.attrs['class'] = 'form-control form-control-sm w-100'
调查和