Django模板for循环(使用for循环打印有限数据)

Question

假设我有 100 部电影。在for循环的帮助下。我们可以打印所有电影。就像下面这样。

在 Django 模板中

{% for movie in movies.object_list %}

    {% endfor %}

但是如果我只需要打印 1 25 50 75 100 我该怎么办 列表中的电影？谢谢

更新:我已经写了这个。还有其他选择吗？

{% for movie in movies.object_list %}
        {% if forloop.counter == 25 %}

            {{ movie }}

        {% endif %}

        {% if forloop.counter == 50 %}

            {{ movie }}

        {% endif %}
        {% if forloop.counter == 75 %}

            {{ movie }}

        {% endif %}
        {% if forloop.counter == 100 %}

            {{ movie }}

        {% endif %}
    {% endfor %}

查看

def movie_sort(request):
categories = Category.objects.all()
language_name = Category.objects.get(id=request.GET.get('language'))
movie_l = Movie.objects.filter(language=request.GET.get('language'),is_active=True)
   # Displaying first row 
for i, v in enumerate(movie_l):
   if i == 0:
       first_row_f = v
   if i == 24:
       first_row_l = v
   if i == 25:
       second_row_f = v
   if i == 49:
       second_row_l = v
   if i == 50:
       third_row_f = v
   if i == 74:
       third_row_l = v
   if i == 75:
       fourth_row_f = v
   if i == 99:
       fourth_row_l = v
################
paginator = Paginator(movie_l, 100) # Show 25 contacts per page

page = request.GET.get('page',1)
try:
    movies = paginator.page(page)
except PageNotAnInteger:
    movies = paginator.page(1)
except EmptyPage:
    movies = paginator.page(paginator.num_pages)
return render_to_response('movie/alphabetic_list.html',locals(),
                          context_instance=RequestContext(request))

更新2:我在 View 中编写了以下函数。唯一的问题是，如果用户单击下一页(分页器)，它会显示以前的数据。意味着这个数据不会改变。有什么建议吗？

  for i, v in enumerate(movie):
   if i == 0:
       first_row_f = v
   if i == 24:
       first_row_l = v
   if i == 25:
       second_row_f = v
   if i == 49:
       second_row_l = v
   if i == 50:
       third_row_f = v
   if i == 74:
       third_row_l = v
   if i == 75:
       fourth_row_f = v
   if i == 99:
       fourth_row_l = v

Answer 1

如果您只是想要每 25 个项目，@Gautam K 就很接近:

{% for movie in movies.object_list|slice:"::25" %}
    {{ movie }}
{% endfor %}

上述解决方案包括切片后缺少的冒号，并且适用于任何大小的列表(Gautam 的解决方案仅适用于 100 项列表)。