Django get latest for each group
WebDec 11, 2024 · 1 Answer Sorted by: 0 The dataset you need can be efficiently built with prefetch_related. userinfo = UserInfo.objects.all ().prefetch_related ("attendancerecord_set") for ui in userinfo: for ar in ui.addendancerecord_set.all (): print (ar) in template code Web[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:
Django get latest for each group
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WebAug 21, 2015 · 1. I have a little problem with getting latest foreign key value in my django app. Here are my two models: class Stock (models.Model): ... class Dividend (models.Model): date = models.DateField ('pay date') stock = models.ForeignKey (Stock, related_name="dividends") class Meta: ordering = ["date"] I would like to get latest … WebMar 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)?
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. WebMay 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
WebOct 7, 2015 · Now, i want to get the problems count for latest test for each name. For example if my table data is. name date problems count foo 10.10.15 50 foo 10.09.15 30 bar 10.07.15 23 foo 10.03.15 54 bar 05.03.15 31 foo 10.01.15 97 Then i would like to get. WebJul 26, 2013 · Imagine we have the Django ORM model Meetup with the following definition: class Meetup (models.Model): language = models.CharField () date = models.DateField (auto_now=True) I'd like to fetch the latest meetup for each language. It would seem you could use Django Aggregates to make this lookup easy:
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 …
leroy josieWebMay 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 ... leroisamoWebAug 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 … leroy merlin la valette toulonWebJan 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') leroy jolleyWebMar 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. leroy merlin leiria onlineWebfrom 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. leroy merlin lublin olimp kontaktWebJan 25, 2024 · It's also easy to get an individual vendor / locale's current value with Costs.objects.filter(vendor_id=1, locale_id=10).latest(). What I'm interested in getting is all of the latest cost values for each vendor / locale combo. So essentially running the latest() function over each combination and getting a list / queryset as a result. leroy johnson mules