Question

如何在基于类的视图中获取分页器信息？我想从get_context_data中调用一个函数，我需要传递当前页面和最后一页。

views.py：

function abcWords(str as string) as long

    dim i as long, arr as variant

    arr = split(str, chr(32))

    for i=lbound(arr) to ubound(arr)
        abcWords = abcWords - int(not isnumeric(arr(i)))
    next i

end function

Answer 1

此代码片段摘自Django源代码views/generic/list.py：

def get_context_data(self, *, object_list=None, **kwargs):
    """Get the context for this view."""
    queryset = object_list if object_list is not None else self.object_list
    page_size = self.get_paginate_by(queryset)
    context_object_name = self.get_context_object_name(queryset)
    if page_size:
        paginator, page, queryset, is_paginated = self.paginate_queryset(queryset, page_size)
        context = {
            'paginator': paginator,  #<--- (1)
            'page_obj': page,        #<--- (2)
            'is_paginated': is_paginated,
            'object_list': queryset
        }
    else:
        context = {
             ....
        }
    if context_object_name is not None:
        context[context_object_name] = queryset
    context.update(kwargs)
    return super().get_context_data(**context)

这意味着您的context中的page_obj和paginator可用。您可以将表达式编写为：

{{ page_obj.previous_page_number }}  (2)

或

{{ paginator.num_pages }}    (1)

您还可以编写：

def get_context_date(self, *args, **kwargs):
    context = super().get_context_data(*args, **kwargs)
    my_function(  )  
    context['page-list'] = pageList(context['page_obj'].number, 
                                    context['paginator'].num_pages)
    return context

在ListView中获取当前页码

1 个答案: