Django 多个查询集组合成分页

Question

我将来自不同模型的 2 个查询集合并到一个列表中，并使用分页显示为单个列表。

问题是列表中的对象根据创建它们的模型通过分页显示。

当从模型创建新对象时，我该如何修复列表。即使最近创建的对象来自不同的模型，它也会显示在最近创建的对象之后。

示例是 .我有一个用户白板和用户评论的组合查询集。如果最近创建了新的白​​板对象，我将创建另一个注释。注释将通过分页显示在白板对象之后，而不是与分页下方的所有其他注释一起显示，因为它位于不同的模型中

我希望您能了解我想要实现的目标:)

我的模块的一部分

class WhiteBoard(models.Model):

    user = models.ForeignKey(User)
    name = models.CharField(max_length=100)
    created = models.DateTimeField(auto_now_add=True)
    picture = models.OneToOneField('Picture',related_name='picture',blank=True,null=True)
    def __unicode__(self):
        return self.name


class Comment(models.Model):
    created = models.DateTimeField(auto_now_add=True)
    user = models.ForeignKey(User)
    body = models.TextField()
    picture = models.ForeignKey(Picture)

View .py

@login_required
def archive(request):
        user = User.objects.get(username=request.user)
        person = Person.objects.get(user=user)
        users = User.objects.filter(pk__in=person.following.all().values_list('user__pk',flat=True))
        picture = whiteboard .objects.filter(user__in=users).order_by("-created")
        comment = Comment.objects.filter(user__in=users).order_by("-created")
        posts= list(chain(picture,comment))
        paginator = Paginator(posts, 5)

        try: page = int(request.GET.get("page", '1'))
        except ValueError: page = 1

        try:
            posts = paginator.page(page)
        except (InvalidPage, EmptyPage):
            posts = paginator.page(paginator.num_pages)
        return render(request,'archive.html',{'picture':picture,'comment':comment,'posts':posts})

存档.html

{% for post in posts.object_list %}

{{post.name }}
<br>
{{post.body}}
{% endfor %}

        <!-- Next/Prev page links  -->
        {% if posts.object_list and posts.paginator.num_pages > 1 %}
        <div class="pagination" style="margin-top: 20px; margin-left: -20px; ">
            <span class="step-links">
                {% if posts.has_previous %}
                        <a href= "?page={{ posts.previous_page_number }}">newer entries &lt;&lt; </a>
                {% endif %}

                    <span class="current">
                    &nbsp;Page {{ posts.number }} of {{ posts.paginator.num_pages }}
                </span>

            {% if posts.has_next %}
                    <a href="?page={{ posts.next_page_number }}"> &gt;&gt; older entries</a>
            {% endif %}
        </span>
    </div>
    {% endif %}

</div>

Answer 1

您只需要sort您的组合查询集列表(按它们共享的 created 属性列出):

from itertools import chain
...
posts = list(
            sorted(
                chain(picture,comment),
                key=lambda objects: objects.created,
                reverse=True  # Optional
            ))
paginator = Paginator(posts, 5)
...

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