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。如下所示。

在Django模板中

{% for movie in movies.object_list %}

    {% endfor %}

但我该怎么办？如果我只打印列表中的1 25 50 75 100电影？感谢

更新：我写过这个。还有另一种选择吗？

{% 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 %}

View

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))

UPDATE2：我在view写了以下函数。如果用户点击下一页（paginator），它显示以前的数据是唯一的问题。意味着这些数据不会改变。任何建议？

  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

如果您只想要每25个项目，@ Gautam K就会关闭：

{% 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，适用于任何大小列表（Gautam的解决方案仅适用于100个项目列表）。

Answer 2

执行此操作的两种基本方法

pagination
MultipleObjectMixin自动完成整个事情（并使用相同的pagination功能）

编辑 - 以下是如何使用分页器的示例。

选项＃1

首先，修改views.py：

from django.core.paginator import Paginator, EmptyPage, PageNotInteger
from myapp.models import Movie

def movies(request):
    movie_list = Movie.objects.all()
    pg = Paginator(movie_list, 50) # Show 50 items per page
    current_page = request.GET.page('page')
    try:
      movies = pg.page(page) # Grab the page from the URL, like /?page=2
    except PageNotAnInteger:
      movies = pg.page(1) # Start from page 1 if no page was passed
    except EmptyPage:
      movies = pg.page(pg.num_pages) # If invalid number pages; show last page

    # You pass the paginator object, not the queryset
    return render_to_response('mytemplate.html',{'movies':movies})

接下来，调整模板：

{% for movie in movies %}
   {{ movie }}
{% endfor %}

{% if movies.has_previous %}
<a href="?page={{ movies.previous_page_number }}">go back</a>
{% endif %}

You are on page {{ movies.number }} of {{ movies.paginator.num_pages }}

{% if movies.has_next %}
<a href="?page={{ movies.next_page_number }}">go forward</a>
{% endif %}

Answer 3

尝试使用django slice

切割列表

例如尝试

根据@JeremyLewis

的建议编辑

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

而不是if语句

我没有检查过代码，因此可能会有效。

Answer 4

您必须加入enumerate(movie.object_list)

    for i, v in enumerate(movie.object_list):
   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

用于循环的Django模板（使用for循环打印有限数据）

4 个答案: