无法使用djangajax从python返回JSON对象

Question

尝试将此ajax插件应用于djangohttps://github.com/yceruto/django-ajax 我已成功使用ajaxPost，但无法应用ajaxGet

在视图.py. 在python中，所有数据都打印得很好。在

    @ajax
    def notify(request):
     notifications = Notification.objects.filter(whom=request.user.profile)
     for acd in notifications:
        print(acd)
        print(acd.choice_afl)
        print(acd.whom)
        print(acd.who_did)
     return {'notifications': notifications}

在html中：

^{pr2}$

如果我警告noty我得到"notifications"，如果我警告acd.choice_afl我得到undefined（或者如果插入到html中就什么都没有），尽管在python中我得到了我需要的结果。怎么了？python对象不转换成JSON对象吗？在

编辑：为了安全起见，我将django对象序列化为json，如下所示。没有变化

data = serializers.serialize("json",notifications)
return {'notifications': data}

编辑2：我观察了JSON的结构

 var jsonPretty = JSON.stringify(JSON.parse(data),null,2); 
$('#the_very_nw').html(jsonPretty);

看起来像这样。所以我错过了“田野”，但它什么也没改变。还是什么也没印出来。如果我返回HttpResponse:

[ { "fields": { "how_much_new_notifications": 0, "number_of_new_notifications": 0, "who_did": 2, "question_answered_object": null, "whom": 1, "choice_afl": "F" }, "model": "blog.notification", "pk": 1 }, { "fields": { "how_much_new_notifications": 0, "number_of_new_notifications": 0, "who_did": 2, "question_answered_object": null, "whom": 1, "choice_afl": "F" }, "model": "blog.notification", "pk": 2 }, { "fields": { "how_much_new_notifications": 0, "number_of_new_notifications": 0, "who_did": 2, "question_answered_object": null, "whom": 1, "choice_afl": "F" }, "model": "blog.notification", "pk": 3 }, { "fields": { "how_much_new_notifications": 0, "number_of_new_notifications": 0, "who_did": 2, "question_answered_object": null, "whom": 1, "choice_afl": "F" }, "model": "blog.notification", "pk": 4 }, { "fields": { "how_much_new_notifications": 0, "number_of_new_notifications": 0, "who_did": 2, "question_answered_object": null, "whom": 1, "choice_afl": "F" }, "model": "blog.notification", "pk": 5 }, { "fields": { "how_much_new_notifications": 0, "number_of_new_notifications": 0, "who_did": 2, "question_answered_object": null, "whom": 1, "choice_afl": "F" }, "model": "blog.notification", "pk": 6 } ]

Answer 1

我认为错误出在你的JS代码中。尝试按如下方式重写循环：

for (var i = 0; i < notifications.length; i++) {
   var acd = notifications[i];
   $('#the_very_nw').html(acd.choice_afl);
}

当前acd只是数组的索引，而不是它的元素（一个很好的解释here）。在