我认为您描述它的方式可以正常工作,因为在幕后我相信 Django 正在做的是使用 SQL LIMIT 来简单地让数据库完成繁重的工作,即整理出哪些数据和多少数据返回。因为数据库已针对执行此类操作进行了优化,所以执行此操作可能是一种合理的方式。
关键可能是保持查询相同,并且正如您所展示的那样,您可以使用相同的视图来做到这一点。视图可以简单地具有一种模式,这是一种改变分页页数的奇特方式。
你可能会得到这样的网址......
# View all "landscape" items in gallery mode starting on page 3
http://www.example.com/gallery/landscape/multi/3
# View the 45th landscape item in singular mode
http://www.example.com/gallery/landscape/single/45
渲染模板时,分页器将提供has_next 和has_previous 方法,让您知道是否可以使用渲染下一个/上一个链接。
这是我对视图的想法,或者类似的东西(这完全未经测试,并写在我的脑海中)......
url(r'gallery/(?P<category>.+)/(?P<mode>.+)/(?P<offset>\d+)$', 'whatever.views.media_gallery'),
def media_gallery(request, category, mode, offset):
"""
Render a media gallery.
category = media item category filter
mode = ( multi | single )
offset = The pagination offset in multi mode or the media ID in single mode
"""
if mode == 'multi':
per_page = 20 # or however many items per page
elif mode == 'single':
per_page = 1
else:
pass # handle this however
# Queryitems
raw_media_items = Media.objects.filter(category=category)
# Setup paginator
paginator = Paginator(raw_media_items, per_page)
try:
# in multi mode offset is the page offset
# in single mode offset is the media ID
page = int(offset)
except:
page = 1
try:
media_items = paginator.page(page)
except (EmptyPage, InvalidPage):
media_items = paginator.page(paginator.num_pages)
if len(paginated_items) == 1:
# Render single view
return render_to_response('gallery/gallery_view.html',
{ 'media_item':media_items[0], 'paginator':paginator },
context_instance=RequestContext(request) )
else:
# Render gallery view
return render_to_response('gallery/gallery_view.html',
{ 'media_items':media_items, 'paginator':paginator },
context_instance=RequestContext(request) )