Python Django问题视图
我有一个Django的一般观点:Python Django问题视图,python,django,django-templates,django-views,django-generic-views,Python,Django,Django Templates,Django Views,Django Generic Views,我有一个Django的一般观点: def post_list(request, page=0, paginate_by=1, **kwargs): page_size = getattr(settings,'BLOG_PAGESIZE', paginate_by) return list_detail.object_list( request, queryset=Post.objects.published(), paginate_by=page_size, pa
def post_list(request, page=0, paginate_by=1, **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
)
post_list.__doc__ = list_detail.object_list.__doc__
我想在今年发送它,以便在模板中使用它。有人能帮我吗?通过via传递本年度的信息:
from datetime import datetime
year = datetime.now().year
您的核心问题是使用通用视图- 与原始代码相比,您更愿意维护或扩展哪些代码
def post_list(request, page=0, paginate_by=1, **kwargs):
page_size = getattr(settings,'BLOG_PAGESIZE', paginate_by)
queryset = Post.objects.published()
paginator = Paginator(queryset, page_size)
objects = paginator.page(page)
#This stuff is only necessary if you aren't trapping page in the URL conf
#page = request.GET.get('page')
#try:
# objects = paginator.page(page)
#except PageNotAnInteger:
# # If page is not an integer, deliver first page.
# objects = paginator.page(1)
#except EmptyPage:
# # If page is out of range (e.g. 9999), deliver last page of results.
# objects = paginator.page(paginator.num_pages)
return render_to_response('mytemplate.html', {"objects": objects, "year": datetime.now().year}, context=RequestContext(request))