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 SQLLIMIT
来简单地让数据库完成整理要返回的数据和返回多少数据的繁重工作。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,您可以使用相同的视图来做到这一点。该视图可以简单地具有一种模式,这是一种更改分页页数的奇特方式。
你最终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 Next/Previous 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,并且在我的脑海中被写下来)......
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) )