Django泛型ListView的分页不存在

2024-04-20 03:51:38 发布

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我是Django的新手,目前我正在尝试将webscraping后端代码集成到Django项目中。当显示我的结果从webscraping,我无法分页我的结果页,这是一个Django列表视图。在我的ListView下设置paginate\u by不起作用。感谢您的帮助。非常感谢。下面是我的一些代码:

视图.py(paginate_by=5不起作用)

class PostListView(ListView):
    model = ResultsTable
    template_name = 'app/results.html'
    paginate_by = 5

    def get_queryset(self):
        return ResultsTable.objects.all()

    def post(self, request, *args, **kwargs):
        self.queryset = self.get_queryset()
        global name
        name = request.POST.get('input')
        for i in range(len(singaporews)):
            t = threading.Thread(target=Webscraper, args=[singaporews[i], name, webdriver.Chrome('/Users/angyangcheng/chromedriver')])
            allthreads.append(t)
            t.start()
        for i in allthreads:
            i.join()
        return super(PostListView, self).get(request, *args, **kwargs)

    def get(self, request, *args, **kwargs):
        return self.self.response_class(
            request=self.request,
            template=self.get_template_names())

    def get_context_data(self, **kwargs):
        context = super(PostListView, self).get_context_data(**kwargs)
        global name
        context = {
            'links' : ResultsTable.objects.all,            
            'name' : name,
            'count' : ResultsTable.objects.all().count()
        }
        return context

我的post方法下的代码主要是我的webscraping逻辑。你知道吗

你知道吗结果.html(分页代码)

   {% if is_paginated %}
<div class="pagination">
    <span class="page-links">
        {% if page_obj.has_previous %}
            <a href="/cars?page={{ page_obj.previous_page_number }}">previous</a>
        {% endif %}
        <span class="page-current">
            Page {{ page_obj.number }} of {{ page_obj.paginator.num_pages }}.
        </span>
        {% if page_obj.has_next %}
            <a href="/cars?page={{ page_obj.next_page_number }}">next</a>
        {% endif %}
    </span>
</div>
   {% endif %}     

Tags: 代码nameselfobjgetreturnrequestdef