Django的通用视图

Question

在我的看法如下功能一个被称为像

def post_list(request, page=0, paginate_by=20, **kwargs): 
    page_size = getattr(settings,'BLOG_PAGESIZE', paginate_by) 
    return list_detail.object_list(
    request, 
    queryset=Post.objects.published(), 
    paginate_by=page_size, 
    page=page, 
    **kwargs 
) 

我需要通过我的字典到页面我该怎么办呢..

在/ usr/mysite的/ Django的/ Django的/视图/通用list_detail.py

def object_list(request, queryset, paginate_by=None, page=None, 
      allow_empty=True, template_name=None, template_loader=loader, 
      extra_context=None, context_processors=None, template_object_name='object', 
      mimetype=None): 
     """ 
     Generic list of objects. 

     Templates: ``<app_label>/<model_name>_list.html`` 
     Context: 
      object_list 
       list of objects 
      is_paginated 
       are the results paginated? 
      results_per_page 
       number of objects per page (if paginated) 
      has_next 
       is there a next page? 
      has_previous 
       is there a prev page? 
      page 
       the current page 
      next 
       the next page 
      previous 
       the previous page 
      pages 
       number of pages, total 
      hits 
       number of objects, total 
      last_on_page 
       the result number of the last of object in the 
       object_list (1-indexed) 
      first_on_page 
       the result number of the first object in the 
       object_list (1-indexed) 
      page_range: 
       A list of the page numbers (1-indexed). 
     """ 
     if extra_context is None: extra_context = {} 
     queryset = queryset._clone() 
     if paginate_by: 
      paginator = Paginator(queryset, paginate_by, allow_empty_first_page=allow_empty) 
      if not page: 
       page = request.GET.get('page', 1) 
      try: 
       page_number = int(page) 
      except ValueError: 
       if page == 'last': 
        page_number = paginator.num_pages 
       else: 
        # Page is not 'last', nor can it be converted to an int. 
        raise Http404 
      try: 
       page_obj = paginator.page(page_number) 
      except InvalidPage: 
       raise Http404 
      c = RequestContext(request, { 
       '%s_list' % template_object_name: page_obj.object_list, 
       'paginator': paginator, 
       'page_obj': page_obj, 

       # Legacy template context stuff. New templates should use page_obj 
       # to access this instead. 
       'is_paginated': page_obj.has_other_pages(), 
       'results_per_page': paginator.per_page, 
       'has_next': page_obj.has_next(), 
       'has_previous': page_obj.has_previous(), 
       'page': page_obj.number, 
       'next': page_obj.next_page_number(), 
       'previous': page_obj.previous_page_number(), 
       'first_on_page': page_obj.start_index(), 
       'last_on_page': page_obj.end_index(), 
       'pages': paginator.num_pages, 
       'hits': paginator.count, 
       'page_range': paginator.page_range, 
      }, context_processors) 
     else: 
      c = RequestContext(request, { 
       '%s_list' % template_object_name: queryset, 
       'paginator': None, 
       'page_obj': None, 
       'is_paginated': False, 
      }, context_processors) 
      if not allow_empty and len(queryset) == 0: 
       raise Http404 
     for key, value in extra_context.items(): 
      if callable(value): 
       c[key] = value() 
      else: 
       c[key] = value 
     if not template_name: 
      model = queryset.model 
      template_name = "%s/%s_list.html" % (model._meta.app_label, model._meta.object_name.lower()) 
     t = template_loader.get_template(template_name) 
     return HttpResponse(t.render(c), mimetype=mimetype) 

Answer 1

您是否尝试过使用extra_context变量Django的通用视图的？

def post_list(request, page=0, paginate_by=20, **kwargs): 
    page_size = getattr(settings,'BLOG_PAGESIZE', paginate_by) 
    return list_detail.object_list(
    request, 
    queryset=Post.objects.published(), 
    paginate_by=page_size, 
    page=page, 
    extra_context={'my':dict}, 
    **kwargs 
)