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Python 无法从Django视图接收到模板的json响应?_Python_Jquery_Json_Django - Fatal编程技术网

Python 无法从Django视图接收到模板的json响应?

Python 无法从Django视图接收到模板的json响应?,python,jquery,json,django,Python,Jquery,Json,Django,我试图将一个json响应从django视图发送到模板,但当我尝试在ajax中console.log响应时,我什么也得不到。我可能做错了什么?我试图将结果数据从视图传递到ajax成功函数。我还注意到一些奇怪的事情,当我在ajax中提到datatype=json时,我在success函数中没有收到任何响应,但当我删除datatype=json时,我在success函数中得到了模板的整个html作为响应。这是为什么 views.py class ChangePassword(View):

我试图将一个json响应从django视图发送到模板,但当我尝试在ajax中console.log响应时,我什么也得不到。我可能做错了什么?我试图将结果数据从视图传递到ajax成功函数。我还注意到一些奇怪的事情,当我在ajax中提到datatype=json时,我在success函数中没有收到任何响应,但当我删除datatype=json时,我在success函数中得到了模板的整个html作为响应。这是为什么

   views.py
    class ChangePassword(View):
        def post(self, request, *args, **kwargs):
            form = PasswordChangeForm(request.POST)

            #current_password = json.loads(get_value.current_password)
            #print ('current password',get_value['current_password'])

            if form.is_valid():
                print("valid form")
                user = CustomUser.objects.get(email=request.user.email)
                current_password = form.cleaned_data['current_password']
                print('current_password',current_password)

                new_password = form.cleaned_data['new_password']
                print('newpassword',new_password)

                if user.check_password(current_password):
                    print('entered')
                    update_session_auth_hash(self.request, self.request.user)  # Important!
                    user.set_password(new_password)
                    user.save()
                    result = {'success': "Succefully reset your password"};

                    result = json.dumps(result)
                    print ('result',result)


                    return render(request, 'change_password.html',context={'result': result})

                else:
                    return render(request, 'change_password.html', {'error': "We were unable to match you old password"
                                                                " with the one we have. <br>"
                                                                "Please ensure you are entering your correct password"
                                                                "then try again."})
            else:
                print("not valid")
                return render(request, 'change_password.html', {'form':form})

        def get(self, request, *args, **kwargs):
            return render(request, 'change_password.html', {})

template
  function changePassword() {
            csrfSetUP()
            new_pass = document.getElementById('new_pass')
            cur_pass = document.getElementById('current_pass')
            if (validpassword(new_pass) && cur_pass!= "") {
                var cpass = $('#current_password').val();
                var npass = $('#new_pass').val();
                var data = {current_password:cpass,new_password:npass}



                 $.ajax({
                        url: '/account/change_password/',
                        type: 'post',
                        data: data,
                        dataType: "json",

                        success: function(json) {
                            console.log(json)
                        }
                    });







            } else {
                $('#change_password').submit()
            }

        }

使用AJAX时,必须使用JSONResponse而不是渲染

如果您需要使用JSON生成一些HTML,那么最好使用render_to_string并将HTML字符串返回到AJAX,smth如下所示: html=render_to_string'ajax/product_details.html',{most_visted:most_visted_prods} 返回HttpResponsehtml 注意:当使用render_to_string时,请记住从AJAX中删除dataType:json,因为返回的不是json,而是字符串。 下面有很多这样的例子,但是看看新的例子

from django.http import JsonResponse
return JsonResponse({'foo':'bar'})