Question

我想呈现两个不同的HTML示例，并将其作为对ajax请求的响应发回。

我认为我有这样的事情：

def getClasses(request):
   User = request.user 
   aircomcode = request.POST.get('aircompany_choice', False)

   working_row = Pr_Aircompany.objects.get(user=User, aircomcode=aircomcode)
   economy_classes = working_row.economy_class
   business_classes = working_row.business_class

   economy = render_to_response('dbmanager/classes.html', {"classes": economy_classes}, content_type="text/html")
   business = render_to_response('dbmanager/classes.html', {"classes": business_classes}, content_type="text/html")

   return JsonResponse({"economy": economy, 
                    "business": business})

有了这个，我得到了错误：

位于0x7f501dc56588的django.http.response.HttpResponse对象不是JSON可序列化的＆＃34;

我该如何完成任务？

在获得响应的js中，我想将接收到的HTML插入到corespoding块中。像这样：

$.ajax({ # ajax-sending user's data to get user's classes
    url: url,
    type: 'post',
    data: {"aircompany_choice": aircompany_choice}, # send selected aircompanies for which to retrieving classes required
    headers: {"X-CSRFToken":csrftoken}, # prevent CSRF attack
}).done (result) ->
    add_booking_classes.find(".economy-classes").children(":nth-child(2)").html(result["economy"])
    add_booking_classes.find(".business-classes").children(":nth-child(2)").html(result["business"])

Answer 1

尝试Django的render_to_string：

economy = render_to_string('dbmanager/classes.html', {"classes": economy_classes})
business = render_to_string('dbmanager/classes.html', {"classes": business_classes})

render_to_string()加载模板，渲染它然后返回结果字符串。然后，您可以将这些结果字符串作为JSON发送。

您的最终代码现在变为：

from django.template.loader import render_to_string

def getClasses(request):
   User = request.user 
   aircomcode = request.POST.get('aircompany_choice', False)

   working_row = Pr_Aircompany.objects.get(user=User, aircomcode=aircomcode)
   economy_classes = working_row.economy_class
   business_classes = working_row.business_class

   economy = render_to_string('dbmanager/classes.html', {"classes": economy_classes})
   business = render_to_string('dbmanager/classes.html', {"classes": business_classes})

   return JsonResponse({"economy": economy, 
                    "business": business})

Answer 2

顾名思义，

render_to_response用于呈现响应。你不想这样做;您想要渲染两个模板，并将它们放入JSON响应中。因此，请使用render_to_string。

Answer 3

您可以在上下文中发送一个，也可以在要渲染的位置发送一个。

如何在Django中使用html作为JSON发送两个变量？

3 个答案: