python之如何在 Django 中将两个带有 html 的变量作为 JSON 发送

我想呈现两个不同的 HTML 示例并将其作为对 ajax 请求的响应发送回去。

我的观点是这样的:

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"

我怎样才能完成我的任务？

在 js 中，当我收到响应时，我想将收到的 HTML 插入到 corespoding block 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"]) 

请您参考如下方法:

试试 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() 加载一个模板，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})