有没有办法让查询尊重输入参数的顺序？
（如果这完全荒谬，请告诉我，这可能是我没有找到任何相关内容的原因。）
这个故事有两个模型 Ranking 和 Artist、Ranking 一般与 Artist 相关（object_id、content_type...整个 shebang）。
我有一个由 Ranking.objects.values_list() 返回的对象列表，按特定字段排序（在我的例子中是 score）。显然，如果我想显示已排名的艺术家列表，我希望它们按相同的顺序排列。我尝试了不同的方法，例如 .filter(pk__in=list)、.in_bulk(list) 等。尝试强制 .values_list( ) 也放入元组中。
他们都获取我的列表：
>>> objects = Ranking.objects.filter(<stuff>).order_by('score')
>>> objects_list = objects.values_list('object_id', flat=True)
>>> objects_list
[8, 1, 2, 15, 14, 3, 13, 31, 16, 5, 4, 7, 32, 9, 37]
并像这样返回它：（
>>> Artist.objects.filter(id__in=objects_list).values_list('id', flat=True)
[7, 32, 3, 8, 4, 2, 31, 9, 37, 13, 16, 1, 5, 15, 14]
为了理智起见，我只是在第二种情况下给出 ID。）
现在，我可以让它工作的唯一方法是创建一个空列表并循环遍历非 values_list() 查询。
for item in objects:
ranked_items.append(item.content_object)
这只会产生 n 个查询，所以我想知道是否有更好的方法。如标签所示，我正在使用 PostgreSQL。
(Please let me know if this is completely absurd, that is probably a reason why I haven't found anything on this.)
This story has two models Ranking and Artist, Ranking is generically related to Artist (object_id, content_type... the whole shebang).
I have a list of objects returned by Ranking.objects.values_list() ordered by a certain field (in my case score). So obviously, if I want to display a list of artists that were ranked, I'd want them in the same order. I've tried different methods, such as .filter(pk__in=list), .in_bulk(list), etc. Tried coercing the result of .values_list() into a tuple too.
They all take my list:
>>> objects = Ranking.objects.filter(<stuff>).order_by('score')
>>> objects_list = objects.values_list('object_id', flat=True)
>>> objects_list
[8, 1, 2, 15, 14, 3, 13, 31, 16, 5, 4, 7, 32, 9, 37]
And return it like so:
>>> Artist.objects.filter(id__in=objects_list).values_list('id', flat=True)
[7, 32, 3, 8, 4, 2, 31, 9, 37, 13, 16, 1, 5, 15, 14]
(I'm just giving the IDs in the second case for sanity's sake.)
Right now the only way I can get this to work is to create an empty list and loop through the non-values_list() query.
for item in objects:
ranked_items.append(item.content_object)
This just produces n queries, so I'm wondering if there's a better way. As shown by the tags, I'm using PostgreSQL.
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这不是 Django 造成的；由于未指定顺序，您的数据库将按此顺序返回它们。
This is not caused by Django; your database is returning them in this order since no ordering is specified.