Question

我目前正在使用以下jquery ajax方法通过POST请求提交数据，并且我在我的控制台（见下文）中成功获得了返回结果，该结果显示了提交给服务器的JSON请求。但是，我无法弄清楚如何在我的Django视图中解析JSON对象。我该怎么写我的视图，这样我才能解析我的JSON对象并获得＆＃34; schedule_name＆＃34;在我的Django视图中执行以下命令。请在下面查看我的观点副本。

Schedule.objects.create（schedule_name = Schedule）

＆＃13;

$.ajax({

type: "POST",
url: "{% url 'addSchedule' building_pk %}",
dataType: "json",
data: {"json_items" : JSON.stringify(Schedule_Info)},
success : function(data) 
  {
    $('.todo-item').val(''); // remove the value from the input
    console.log(data); // log the returned json to the console
    console.log("success"); // another sanity check
    alert("Sucess");
     
  },

＆＃13;

提交请求后在控制台中输出JSON

＆＃13;

json-items: "{

"nameOfSchedule":{"schedule_name":"Schedule"},

"Rooms":[
	{"room_rank":"1","room_name":"Room 101 "},
	{"room_rank":"2","room_name":"Room 102 "},
	{"room_rank":"3","room_name":"Room 103 "},
	{"room_rank":"4","room_name":"Room 104 "}
	],

"Users":[
	{"user_name":"test1@yahoo.com"},
	{"user_name":"test2@yahoo.com"}
	]
}"

＆＃13;

Django View

def addSchedule(request, building_id):

    building_pk = building_id

    b = Building.objects.get(pk=building_pk)
    floor_query = b.floor_set.all()
    master_query = floor_query.prefetch_related('room_set').all()

    if request.is_ajax() and request.POST:
        data = request.POST

    ### Input the schedule name in the datase by parsing the JSON object

        return HttpResponse(json.dumps(data),content_type="application/json")

    else:
        return render(request, 'scheduler/addSchedule.html', {'building_pk' : building_pk,
                                                        'building_query': master_query

                                                          })

Answer 1

您可以使用JsonResponse：

return JsonResponse({'this': "will be converted to json"});

有关详细信息，请参阅https://docs.djangoproject.com/el/1.10/ref/request-response/

Answer 2

我通过对Django进行以下更改来解决问题

Django view

data = json.loads(request.POST.get('json_items'))

name = data['nameOfSchedule']['schedule_name']
Schedule.objects.create(schedule_name = name)

如何从Ajax POST中解析Django View中的JSON对象

2 个答案: