2024 Django get latest for each group

Django get latest for each group

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WebSep 10, 2024 · A simple way to do this in a single query is to annotate each photo with the latest date for the related person and then filter by the annotation. This should return all desired PersonPhoto in a queryset. from django.db.models import Max, F PersonPhoto.objects.annotate ( latest=Max ('person__personphoto__date_captured') … 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:

Django ORM: Group by and Max - Stack Overflow

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 31, 2024 · 1 Answer. Sorted by: 2. You may try this: MyModel.objects.filter ( #prepare subquery which has max dates per month #and limit resultset only to those dates date__in = Subquery ( MyModel.objects.annotate ( truncated_to_month = TruncMonth ('date') #used to group by truncated to month date ).values ( 'truncated_to_month' ).annotate ( … everything i know about women

python - Queryset select latest record for group - Stack Overflow

WebMay 19, 2024 · yes, you can go ahead and test it. and print and check the queryset returned. I have a similar use case (considering the same model in the question), instead I had to fetch the latest record for each seller on each date. So I did this :- latest_sale = Sales.objects.order_by ("Saler", "-Date").distinct ("Saler") – Abhijeet Anand Shah. 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') 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 … brown soft gel pills

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[Answered]-Django Query Get Last Record with a Group By-django

WebDjango 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 … 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 … everything i know i learned in kindergartenWebMar 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 … everything i know i learned from my mom

"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. " - Django get latest for each group

Django get latest for each group

Get last record for list of ids Django - Stack Overflow

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 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.

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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: 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.

WebHow do I write a Django ORM query that: groups the Requests by user, filters the Requests based on req_text, and also, select the max id of the resulting result set. So for each user, I will return one row which matches the filter condition and also has the greatest id. django django-orm Share Follow asked Aug 4, 2011 at 11:02 jeffreyveon 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 ...

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

WebGROUP BY and Select MAX from each group in Django, 2 queries Raw gistfile1.txt ''' given a Model with: category = models.CharField (max_length=32, choices=CATEGORY_CHOICES) pubdate = models.DateTimeField (default=datetime.now) Fetch the item from each category with the latest pubdate. '''

brown softball glovesWebOct 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. brown softball schedule 2023WebDec 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 everything i know paul jarvisWebJan 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. everything ikonWebJul 5, 2024 · I want to get the latest score record for each student. I have tried: Score .objects ( 'student' .annotate (latest_date= Max 'date' )) Copy. and: Score … everything i learned in lifeWebMar 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)? everything illegalWebDjango 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 … everything illuminated sountrack