Question

将django-secretballot与我的应用程序“flavor”集成后，此500状态错误不是CSRFtoken问题。 FireBug网络选项卡显示

POST：

content_type    flavor.Flavor
object_id   5
vote    1
Source
content_type=flavor.Flavor&object_id=5&vote=1

响应：

TypeError at /flavor/vote/
vote() missing 3 required positional arguments: 'content_type', 'object_id', and 'vote'

你能发现我的flavor_vote.js中的任何错误吗？为什么不能将这3个必需的位置参数传递给vote（）函数？

$(document).ready(function() {

  var csrftoken = getCookie('csrftoken');

  $.ajaxSetup({
      beforeSend: function(xhr, settings) {
          if (!csrfSafeMethod(settings.type) && !this.crossDomain) {
              xhr.setRequestHeader("X-CSRFToken", csrftoken);
          }
      },
      error: function(xhr, textStatus, error) {
          console.log(error);
      }
  });

  $(".vote").click(function () {
    var div = $(this);
    var obj_id = $(this).closest('.flavor').attr('flavor-id');
    var vote = 1;
    if ($(this).hasClass('voted')) {
        var vote = 0;
    }

   console.log(div);
    $.ajax({
        url: '/flavor/vote/',
        method: 'post',
        data: {
            'content_type': 'flavor.flavor',
            'object_id': obj_id,
            'vote': vote
        },
        success: function (data) {
            alert('Successfully called');
            $(div).toggleClass('btn-success voted');
            console.log(data);
        },
        error: function(data) {
            console.log(data);
        }
    });
  });
});

function getCookie(name) {
    var cookieValue = null;
    if (document.cookie && document.cookie != '') {
        var cookies = document.cookie.split(';');
        for (var i = 0; i < cookies.length; i++) {
            var cookie = jQuery.trim(cookies[i]);
            // Does this cookie string begin with the name we want?
            if (cookie.substring(0, name.length + 1) == (name + '=')) {
                cookieValue = decodeURIComponent(cookie.substring(name.length + 1));
                break;
            }
        }
    }
    return cookieValue;
}

function csrfSafeMethod(method) {
    // these HTTP methods do not require CSRF protection
    return (/^(GET|HEAD|OPTIONS|TRACE)$/.test(method));
}

========其他相关信息================

我的models.py有Flavor类，可以投票

secretballot.enable_voting_on(Story)

我的ajax调用需要将参数传递给https://github.com/jamesturk/django-secretballot/blob/master/secretballot/views.py中的views.py中的vote()函数

def vote(request, content_type, object_id, vote, can_vote_test=None,
         redirect_url=None, template_name=None, template_loader=loader,
         extra_context=None, context_processors=None, mimetype=None):

urls.py有

url(r'^flavor/vote/$', 'secretballot.views.vote', name='flavor_vote'),

django template flavor_detail.html

<div class="caption flavor" flavor-id="{{ flavor.id }}">
    <div class="btn btn-default vote">
       <span class="glyphicon glyphicon-thumbs-up"></span>
       Vote
    </div>
</div>

Answer 1

您的vote函数应该只使用一个参数request：

def vote(request):

通过AJAX传递给视图的变量将出现在request.POST dict中。它们不会作为args或kwargs传递给视图函数。

def vote(request):
    content_type = request.POST['content_type']
    vote = request.POST['vote']
    object_id = request.POST['object_id']
    ...

Answer 2

需要改变两件事......

第一

在urls.py中添加对位置参数的支持

url(r'^flavor/vote/(?P<content_type>)/(?P<object_id>)/(?P<vote>)', 'secretballot.views.vote', name='flavor_vote'),

第二

像这样更改你的ajax请求。

type = "some type"; // fetch it somehow.
$.ajax({
        url: '/flavor/vote/'+type+'/'+obj_id+'/'+ vote,
        method: 'post',
        data: { //not needed now as the url contains it.
            'content_type': 'flavor.flavor',
            'object_id': obj_id,
            'vote': vote
        },
        success: function (data) {
            alert('Successfully called');
            $(div).toggleClass('btn-success voted');
            console.log(data);
        },
        error: function(data) {
            console.log(data);
        }
    });

Ajax POST调用Django的多视图视图[Response 500]

2 个答案: