django-tables2：在渲染方法中使用请求用户

Question

我正在使用 djnago-tables2 来显示来自我的事件模型的数据。我需要让当前请求用户进入 table 的渲染方法（进一步调用模型函数来获取当前用户的响应）。

我该怎么做？

我的模型

class Event(models.Model):

    name=models.CharField(max_length=254)
    start=models.DateTimeField()
    end=models.DateTimeField()
    invited_user=models.ManyToManyField(User,blank=True)

    def getResponseFromUser(self, user):
        try:
            return Response.objects.filter(event=self, user=user).order_by('-pk')[0]
        except Exception:
            return None

class Response(models.Model):
    user=models.ForeignKey(User)
    event=models.ForeignKey(Event)
    timestamp=models.DateTimeField(auto_now_add=True)

    SEEN=0
    YES=1
    PROBABLY_YES=2
    PROBABLY_NO=3
    NO=4

    RESPONSE=(
        (SEEN, _('seen but undecided')),
        (YES, _('Yes')),
        (PROBABLY_YES, _('probably yes')),
        (PROBABLY_NO, _('probably no')),
        (NO, _('No')),
    )
    response=models.IntegerField( choices=RESPONSE, default=SEEN)

    @property
    def getResponseText(self):
        return self.RESPONSE[self.response][1]

Table

class EventTable(tables.Table):

    requestUser=None # to be set from request (externally)

    eventResponse = tables.Column(empty_values=(), verbose_name="Response", orderable=False)

    class Meta:
        model = Event
        # add class="paleblue" to <table> tag
        attrs = {"class": "table"}
        exclude=['id', 'user_can_invite', 'description', 'organizer']

    def render_eventResponse(self, record):
        responseObject=record.getResponseFromUser(self.requestUser)
        if not responseObject:
            return ''
        else:
            return responseObject.getResponseText

查看

class PersonalEventTableView(SingleTableView):
    model = Event
    table_class = EventTable
    template_name = 'list_events.html'
    paginate_by = 50

在 urls.py 中，视图使用 PersonalEventTableView.as_view()

呈现

有人能帮忙吗？

Answer 1

你可以在这里使用这个插件：https://github.com/Alir3z4/django-crequest

这样使用：

def render_eventResponse(self, record):
    from crequest.middleware import CrequestMiddleware
    current_request = CrequestMiddleware.get_request()
    user = current_request.user
    responseObject=record.getResponseFromUser(user)
    if not responseObject:
        return ''
    else:
        return responseObject.getResponseText

Answer 2

您可以非常简单地将请求添加到 table 的上下文中。

class PersonalEventTableView(SingleTableView):
    model = Event
    table_class = EventTable
    template_name = 'list_events.html'
    paginate_by = 50

    def get_context_data(self, **kwargs):
        """ Override SingleTableView class method """
        context = super(SingleTableMixin, self).get_context_data(**kwargs)
        table = self.get_table(**self.get_table_kwargs())
        
        # add the request directly to your table
        table.request = self.request
        context[self.get_context_table_name(table)] = table
        return context

现在可以在您的 Table class 中使用它。

class 事件Table(tables.Table):

eventResponse = tables.Column(empty_values=(), verbose_name="Response", orderable=False)

class Meta:
    model = Event
    attrs = {"class": "table"}
    exclude=['id', 'user_can_invite', 'description', 'organizer']

def render_eventResponse(self, record):
    # self.request.user is available
    responseObject=record.getResponseFromUser(self.request.user)
    return responseObject.getResponseText if responseObject else ""

django-tables2：在渲染方法中使用请求用户

django-tables2: use request user in render method

django

django-models

django-tables2