添加对象时的 Django 分页

Django pagination while objects are being added

我有一个网站显示的照片总是被添加,人们看到主页上的页面之间存在重复(最后添加的照片)

我不完全确定如何解决这个问题,但这基本上是正在发生的事情:

解决这个问题的最佳方法是什么?我想我可能需要以某种方式在用户会话中缓存查询集?我真的不太了解缓存,所以一步一步的解释将是无价的

这是获取新页面图像的函数:

def get_images_paginated(query, origins, page_num):
    args = None
    queryset = Image.objects.all().exclude(hidden=True).exclude(tags__isnull=True)
    per_page = 20
    page_num = int(page_num)
    if origins:
        origins = [Q(origin=origin) for origin in origins]
        args = reduce(operator.or_, origins)
        queryset = queryset.filter(args)        


    if query:
        images = watson.filter(queryset, query)
    else:
        images = watson.filter(queryset, query).order_by('-id')
    amount = images.count()
    images = images.prefetch_related('tags')[(per_page*page_num)-per_page:per_page*page_num]

    return images, amount

使用函数的视图:

def get_images_ajax(request):
    if not request.is_ajax():
        return render(request, 'home.html')

    query = request.POST.get('query')
    origins = request.POST.getlist('origin')
    page_num = request.POST.get('page')

    images, amount = get_images_paginated(query, origins, page_num)
    pages = int(math.ceil(amount / 20))
    if int(page_num) >= pages:
        last_page = True;
    else:
        last_page = False;
    context = {
        'images':images,
        'last_page':last_page,
    }

    return render(request, '_images.html', context)

您可以采取的一种方法是在 AJAX 请求中发送客户端当前拥有的最旧 ID(即当前列表中最后一项的 ID),然后确保您只查询旧 ID。

所以get_images_paginated修改如下:

def get_images_paginated(query, origins, page_num, last_id=None):
    args = None
    queryset = Image.objects.all().exclude(hidden=True).exclude(tags__isnull=True)
    if last_id is not None:
        queryset = queryset.filter(id__lt=last_id)
    ...

您需要在 AJAX 请求中发送最后一个 ID,并将其从您的视图函数传递给 get_images_paginated:

def get_images_ajax(request):
    if not request.is_ajax():
        return render(request, 'home.html')

    query = request.POST.get('query')
    origins = request.POST.getlist('origin')
    page_num = request.POST.get('page')
    # Get last ID. Note you probably need to do some type casting here.
    last_id = request.POST.get('last_id', None)

    images, amount = get_images_paginated(query, origins, page_num, last_id)
    ...

正如@doniyor 所说,您应该结合此逻辑使用 Django 的内置分页。