Django 升序和降序

Question

假设，我想按Votes、Views 和Comments 的总数订购Threads。但是他们order模型是从Thread模型实例化的。

这是我的models.py

class Thread(models.Model):
    ....
    rating = RatingField(can_change_vote=True)

    def get_comments(self):
        return Comment.objects.filter(thread__pk=self.pk)

    def get_visitors(self):
        return Visitor.objects.filter(thread__pk=self.pk)

    def get_rating_frequence(self):
        return self.rating.get_difference()

class Comment(models.Model):
    thread = models.ForeignKey(Thread, ...)
    ....

class Visitor(models.Model):
    thread = models.ForeignKey(Thread, ...)
    ....

和Vote的模型使用django updown

更多：https://github.com/weluse/django-updown/blob/master/updown/models.py#L20

class Vote(models.Model):
    content_type = models.ForeignKey(ContentType, related_name="updown_votes")
    object_id = models.PositiveIntegerField()
    ....

这是我的views.py

def threads(request):
    ....
    template_name = 'foo/bar/threads.html'
    threads = Thread.objects.published()

    get_sorted_ratings = request.GET.get('sr', '')
    get_sorted_visitors = request.GET.get('sv', '')
    get_sorted_comments = request.GET.get('sc', '')

    context = {
        'threads': threads,
        ....
    }
    return render(request, template_name, context)

问题是，我如何根据Thread 型号订购多个型号？ 在我的测试中：

>>> from django.db.models import (Q, Count, Sum)
>>> from myapp.models import (Thread, Comment, Visitor)
>>> t = Thread.objects.published()
>>> Visitor.objects.filter(thread__in=t).values('thread').annotate(visit=Count('thread__id')).order_by('-visit')
<QuerySet [{'thread': 3, 'visit': 8}, {'thread': 10, 'visit': 8}, {'thread': 9, 'visit': 7}, {'thread': 48, 'visit': 1}, {'thread': 57, 'visit': 1}, {'thread': 79, 'visit': 1}, {'thread': 103, 'visit': 1}, {'thread': 104, 'visit': 1}, {'thread': 132, 'visit': 1}, {'thread': 178, 'visit': 1}, {'thread': 216, 'visit': 1}, {'thread': 227, 'visit': 1}, {'thread': 267, 'visit': 1}, {'thread': 292, 'visit': 1}, {'thread': 300, 'visit': 1}, {'thread': 305, 'visit': 1}, {'thread': 1201, 'visit': 1}, {'thread': 1252, 'visit': 1}, {'thread': 1265, 'visit': 1}, {'thread': 1929, 'visit': 1}]>
>>> 
>>> Comment.objects.filter(thread__in=t).values('thread').annotate(total=Count('thread__id')).order_by('-total')
<QuerySet [{'total': 9, 'thread': 10}, {'total': 7, 'thread': 9}, {'total': 3, 'thread': 3}, {'total': 3, 'thread': 213}, {'total': 2, 'thread': 35}, {'total': 2, 'thread': 47}, {'total': 2, 'thread': 104}, {'total': 2, 'thread': 187}, {'total': 2, 'thread': 233}, {'total': 2, 'thread': 235}, {'total': 2, 'thread': 304}, {'total': 1, 'thread': 34}, {'total': 1, 'thread': 68}, {'total': 1, 'thread': 82}, {'total': 1, 'thread': 95}, {'total': 1, 'thread': 137}, {'total': 1, 'thread': 216}, {'total': 1, 'thread': 231}, {'total': 1, 'thread': 244}, {'total': 1, 'thread': 253}, '...(remaining elements truncated)...']>
>>> 
>>> # rating likes, id=thread.id, total=total rating
>>> t.values('id').annotate(total=Sum('rating_likes')).order_by('-total')
<QuerySet [{'total': 3, 'id': 10}, {'total': 2, 'id': 104}, {'total': 2, 'id': 233}, {'total': 2, 'id': 235}, {'total': 2, 'id': 304}, {'total': 1, 'id': 3}, {'total': 1, 'id': 9}, {'total': 1, 'id': 34}, {'total': 1, 'id': 35}, {'total': 1, 'id': 47}, {'total': 1, 'id': 68}, {'total': 1, 'id': 82}, {'total': 1, 'id': 95}, {'total': 1, 'id': 137}, {'total': 1, 'id': 187}, {'total': 1, 'id': 213}, {'total': 1, 'id': 216}, {'total': 1, 'id': 231}, {'total': 1, 'id': 244}, {'total': 1, 'id': 253}, '...(remaining elements truncated)...']>
>>> 
>>> # rating dislikes, id=thread.id, total=total rating
>>> t.values('id').annotate(total=Sum('rating_dislikes')).order_by('-total')
<QuerySet [{'total': 2, 'id': 3}, {'total': 1, 'id': 9}, {'total': 1, 'id': 10}, {'total': 1, 'id': 35}, {'total': 1, 'id': 47}, {'total': 1, 'id': 187}, {'total': 0, 'id': 29}, {'total': 0, 'id': 34}, {'total': 0, 'id': 42}, {'total': 0, 'id': 45}, {'total': 0, 'id': 48}, {'total': 0, 'id': 51}, {'total': 0, 'id': 53}, {'total': 0, 'id': 57}, {'total': 0, 'id': 68}, {'total': 0, 'id': 72}, {'total': 0, 'id': 75}, {'total': 0, 'id': 76}, {'total': 0, 'id': 77}, {'total': 0, 'id': 79}, '...(remaining elements truncated)...']>
>>>
>>> Thread.objects.get(pk=3).rating.dislikes
2
>>> Thread.objects.get(pk=9).rating.dislikes
1
>>>

---

最后，我更新了它： 我的代码运行良好，正如@prakhar-trivedi 在https://*.com/a/41775990/6396981 所建议的那样

threads = Thread.objects.published()

def sorted_threads(values, order_by='-total', next_order=None):
    """
    return sorted threads from values by `order_by`
    :param `values`, an example:
        values = Comment.objects.filter(thread__in=threads)
                        .values('thread').annotate(total=Count('thread__id'))
    """
    if next_order is not None:
        top_values = values.order_by(order_by).order_by(next_order)
    else:
        top_values = values.order_by(order_by)
    list_pk_threads = [pk['thread'] for pk in top_values]
    filter_threads = list(threads.filter(pk__in=list_pk_threads))
    return sorted(filter_threads, key=lambda i: list_pk_threads.index(i.pk))

对于在此gist 找到的完整内容。

但是，我就是完全不知道怎么组合。

例如： 从votes=1、views=1、comments=1 订购。 结果是；由 votes 和 views 热门排序的 Thread Objects来自 cmets 的流行。

另一个例子：从votes=1views=1订购。 此条件的输出应该返回 votes 仅从热门排序且 viewsThread Objects /em>。

1 作为True 和-1 作为False

Answer 1

你可以使用 order_by('-thread')，像这样：

Comment.objects.filter(thread__in=t).values('thread').annotate(total=Count('thread__id')).order_by('-thread')

如果你想通过thread模型的rating排序记录，你可以在thread key上使用双下划线“__”进行外键查找， 像这样：

# rating likes

Comment.objects.filter(thread__in=t).values('thread').annotate(total=Count('thread__id')).order_by('thread__rating_likes')

根据要求在所有查询中使用它。