[Django]-Django query get last n records

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]

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 !!

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()

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.

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)]