Django get latest for each group
WebApr 1, 2015 · If you happen to be using PostGreSQL, you can use Django's interface to DISTINCT ON: recent_cakes = Cake.objects.order_by ('bakery__id', '-baked_at').distinct ('bakery__id') As the docs say, you must order by the same fields that you distinct on. As Simon pointed out below, if you want to do additional sorting, you'll have to do it in … WebJan 24, 2024 · To signify this, in Django positional arguments to .distinct () can be passed only in Postgresql ). In Django we can do this with QuerySet like this: Portfolio.objects.order_by ().order_by ( 'code', # first, cause we want to group by this value '-created' # descending order, latest / max will be first ).distinct ('code')
Django get latest for each group
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WebDec 16, 2024 · Get top n records for each group with Django queryset Ask Question Asked 2 years, 3 months ago Modified 2 years, 3 months ago Viewed 1k times 1 I have a model like the following Table, create table `mytable` ( `person` varchar (10), `groupname` int, `age` int ); And I want to get the 2 oldest people from each group. WebMar 19, 2024 · Django ORM group by, and find latest item of each group (window functions) class Cake (models.Model): baked_on = models.DateTimeField (auto_now_add=True) cake_name = models.CharField (max_length=20) Now, there are multiple Cake s baked on the same day, and I need a query that will return me a monthly …
Webfrom django.db.models import Q group_dict = Model.objects.values ('business_id').annotate (max_date=Max ('date')).order_by () params = Q () for obj in group_dict: params = (Q (business_id__exact=obj ['business_id']) & Q (date=obj ['max_date'])) qs = Model.objects.filter (params) This link can help u. WebI can use a queryset to return the latest update time for each location: latest_updates = Locations.objects.values ('location').annotate (max_date=Max ('update_time')).order_by ('location') but this only returns the location and max update_time when I'm looking for the entire row - num_01, num_02, num_03.
Web1 looks like you need to add empty order_by () by the default-ordering-or-order-by in the result query can be added extra fields to the group by. try it: Market.objects.values ('slug').annotate (Max ('active')).order_by () Share Improve this answer Follow answered Oct 23, 2024 at 21:42 Brown Bear 19.4k 10 53 75 Add a comment Your Answer WebMay 11, 2024 · obj = Model.objects.filter (testfield=12).order_by ('id').latest ('id') Filter based on what field you need - in this case is testfield. Model.objects.filter (testfield=12) In order to get the latest record, first you need to sort the queryset. So that it knows to return the last record based on a criteria. Now, order the results base on your ...
WebMar 1, 2024 · We can select top n per group with help of Subquery. Firstly, let's get top n Purchases per customer top_n_purchases_per_customer = Purchases.objects.filter ( customer=OuterRef ('customer') ).order_by ('-field_of_interest') [:10] Next, we can select Purchases with matching ids from the top 10 per each customer.
WebAug 14, 2024 · You can use slice operator to limit the query to a number of records. For example, limit to 10 latest records: UserData.objects.filter (user_id__in=user_list, date_created__lte=start_date) [:10]. Django querysets are lazy so this won't query all records and then slice it - it will only query 10. – vinkomlacic Aug 16, 2024 at 9:39 Add a … premier radiology wisconsinscotrail free wifiWebDjango ORM - Get latest record for group; Filtering Django Query by the Record with the Maximum Column Value; Django Raw Query: Making Count query with group BY … premier radiology white bridge road nashvilleWebMar 29, 2024 · 1 Answer. for each in People.objects.all (): score = models.CharField (verbose_name=each.title, max_length=4) Firstly even if the loop works there would only be one score field in the end. Next a Django model is the reflection of a Database table. Ever heard of a table having an arbitrary number of columns (atleast a good normalized table)? premier range cooker hood sparesWebMay 31, 2024 · Last record of each id ?! each record has only one unique id. But if you mean you want the object with the greatest id I think there are 2 ways to get the last object: queryset = model.objects.filter (id__in= [1, 5, 7]).order_by ('id').last () Or you can do this: queryset = model.objects.filter (id__in= [1, 5, 7]).latest ('id') Share premier ranch properties new mexicoWebDjango orm get latest for each group. I am using Django 1.6 with Mysql. class Student (models.Model): username = models.CharField (max_length=200, unique = True) class … premier ranch texasWeb[Answered]-Django Query Get Last Record with a Group By-django score:-1 Instead .order_by ('-month') [:1] it's better to use .order_by ('month').last () or .order_by ('-month').first () (or earliest / latest for dates). Of course when grouping you can use order_by: scotrail friends and family