Lets say I am implementing a message stream, records sort by ID ascending order, each fetch request, I want to only allow to fetch the most latest 10 records.
I tried:
Messages.objects.filter(since=since)[:-10] And I had an error saying Negative Indexing is not supported.
My current work around is sort ID descending order, and then run:
Messages.objects.filter(since=since)[:10] But this requires the front end to reverse the order again.
My question is, is there a elegant way to do it?
6 Answers
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) 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 !!
2I 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)] 
