Question

我在一般视图中过滤了page_obj中的结果，只显示使用与django-cms（http://www.django-cms.org/en/documentation/2.0/i18n/）当前设置的语言相同的语言发布的条目。

这样可以正常工作，但是添加对Django分页（http://docs.djangoproject.com/en/1.2/topics/pagination/）的支持会导致过滤后的结果仍然被计算在内。所以，例如当英语有三个结果时，从总共十个结果和分页设置为2，我将获得5个结果页面，其中大多数当然是空白的，因为其余七个的过滤是在模板。

我可以使用模板标记修改Django Paginator以使用模板中的过滤器，还是必须重建我的视图？如果是这样，我该怎么做呢？

相关代码：

managers.py

def update_queryset(view, queryset, queryset_parameter='queryset'):
    '''Decorator around views based on a queryset passed in parameter, which will force the update of the queryset before executing the view.
    Related to issue: http://code.djangoproject.com/ticket/8378'''
    def wrap(*args, **kwargs):
        #Regenerate the queryset before passing it to the view.
        kwargs[queryset_parameter] = queryset()
        return view(*args, **kwargs)
    return wrap

视图/ entries.py

from django.views.generic.list_detail import object_list
from cmsplugin_publisher.models import Entry
from cmsplugin_publisher.managers import update_queryset

entry_index = update_queryset(object_list, Entry.published.all)

的URL / entries.py

from django.conf.urls.defaults import *
from cmsplugin_publisher.models import Entry
from cmsplugin_publisher.settings import PAGINATION, ALLOW_EMPTY, ALLOW_FUTURE

entry_conf_list = {'queryset': Entry.published.all(), 'paginate_by': PAGINATION}

entry_conf = { 'queryset': Entry.published.all(),}

entry_conf_detail = entry_conf.copy()
entry_conf_detail['queryset'] = Entry.objects.all()

urlpatterns = patterns('cmsplugin_publisher.views.entries',
    url(r'^$', 'entry_index', entry_conf_list, name='cmsplugin_publisher_entry_archive_index'),
    url(r'^(?P<page>[0-9]+)/$', 'entry_index', entry_conf_list, name='cmsplugin_publisher_entry_archive_index_paginated'),
)

urlpatterns += patterns('django.views.generic.list_detail',
    url(r'^(?P<slug>[-\w]+)/$', 'object_detail', entry_conf_detail, name='cmsplugin_publisher_entry_detail'),
)

在entry_list.html中

{% block content %}
    {% for object in object_list %}
      {% ifequal object.language current_language %}
        ..
      {% endifequal %}
    {% endfor %}
    {% if is_paginated %}
    <ul id="pagination">
    {% if page_obj.has_previous %}
        {% ifnotequal page_obj.start_index 1 %}<li><a href="../" title="{% trans 'View Latest Entries' %}">{% trans 'Latest Entries' %}</a></li>{% endifnotequal %}
        {% ifequal page_obj.previous_page_number 1 %}{% endifequal %}
        {% ifnotequal page_obj.previous_page_number 1 %}
            <li><a href="../{{ page_obj.previous_page_number }}/" title="{% trans 'View Earlier Entries' %}">{% trans 'Earlier Entries' %}</a></li>
        {% endifnotequal %}
    {% else%}
    {% endif %}
    <li>{% trans 'Page' %} {{ page_obj.start_index }} {% trans 'of' %} {{ paginator.num_pages }} {% trans 'Entries' %}</li>
    {% if page_obj.has_next %}
        {% ifnotequal page_obj.start_index 1 %}
            <li><a href="../{{ page_obj.next_page_number }}/" title="{% trans 'View Older Entries' %}">{% trans 'Older Entries' %}</a></li>
        {% endifnotequal %}
        {% ifequal page_obj.start_index 1 %}
        <li><a href="{{ page_obj.next_page_number }}/" title="{% trans 'View Older Entries' %}">{% trans 'Older Entries' %}</a></li>
        {% endifequal %}
    {% else%}
    {% endif %}
    </ul>
{% endif %}

我很感激你们可以在这里找到最好的解决方案。

Answer 1

最终，解决方案是重建视图。在这种情况下需要进行广泛的重建。

故事的道德：不要过滤模板！

在模板中筛选Django paginator的结果

1 个答案: