Django 查询获取最后 n 条记录

假设我正在实现一个消息流，记录按 ID 升序排序，每次获取请求，我只允许获取最新的10条记录。

我试过:

Messages.objects.filter(since=since)[:-10]

我在说不支持负索引时犯了一个错误。

我当前的工作是按降序排序 ID，然后运行:

Messages.objects.filter(since=since)[:10]

但这需要前端再次反转顺序。

我的问题是，有没有一种优雅的方式？

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最佳答案

You can pass your queryset to reversed:

last_ten = Messages.objects.filter(since=since).order_by('-id')[:10]
last_ten_in_ascending_order = reversed(last_ten)
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Or use [::-1] instead of reversed:

last_ten = Messages.objects.filter(since=since).order_by('-id')[:10][::-1]

@ron_g:

Above solution uses list comprehension twice (first to slice to 10, then to reverse). You can do it as one operation, it's 9x faster.

last_ten = Messages.objects.filter(since=since).order_by('-id')[:10:-1]
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If you want last X records sorted in descending order by id , Then I don't think you need since filter

last_ten = Messages.objects.all().order_by('-id')[:10]

Using -id will sort in descending order. Hope this was helpful !!

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In response to ron_g's comment in Omid Raha's answer:

Consider the case where there are less than 10 records in the list:

Note that this approach

list[:10:-1]

will return an empty list, in contrast to

list[:10][::-1]

which will return all the records in the list, if the total records are less than 10.

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I wanted to retrieve the last 25 messages and solved this problem in the following way

#models.py


class Meta:
ordering = ['created']

#views.py


message1 = []
for message in pm_messages.objects.filter(room=room_name).reverse()[0:25]:
message1.append(message)
messagess = message1.reverse()
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you can convert the queryset to a list and select the last n elements normally

messages = list(Messages.objects.filter(since=since))
messages = oldMessageis[-n:len(oldMessageis)]

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