Re: Django ORM generate inefficient SQL

I have created a ticket - https://code.djangoproject.com/ticket/23383 Should I try to proceed my investigations and create a patch? пятница, 29 августа 2014 г., 11:50:01 UTC+6 пользователь Simon Charette написал: > > I think you should create a single issue >

Re: Django ORM generate inefficient SQL

I think you should create a single issue pointing to the fact that only selected columns (either through values() or only()) should be grouped by, regardless of the backend. I suggest you include your initial post body in the report. Thanks! Le

Re: Django ORM generate inefficient SQL

Yes, I guess, in this case it should be able to group only by `auth_group`.`id` for each database backend. Should I create ticket(s) about the issue on MySQL and/or the issue of columns in "group by" statement, or would you like to do it yourself? четверг, 28 августа 2014 г., 20:36:19 UTC+6

Re: Django ORM generate inefficient SQL

The issue on MySQL should be tracked in another ticket I guess -- you're right that it should be able to GROUP only by `auth_group`.`id`. Le jeudi 28 août 2014 07:28:18 UTC-4, Alex Lebedev a écrit : > > Thanks for the answer! Yes, but this problem occurs regardless of > database backend (I

Re: Django ORM generate inefficient SQL

Thanks for the answer! Yes, but this problem occurs regardless of database backend (I tested it on PostgreSQL and MySQL) среда, 27 августа 2014 г., 12:37:00 UTC+6 пользователь Simon Charette написал: > > This is already tracked in #19259 > . > > Le

Re: Django ORM generate inefficient SQL

This is already tracked in #19259 . Le jeudi 21 août 2014 06:49:41 UTC-4, Alex Lebedev a écrit : > > Hi, guys! > > I have encountered a problem. When I use the following code in django > shell: > > from django.contrib.auth.models import Group >

Django ORM generate inefficient SQL

Hi, guys! I have encountered a problem. When I use the following code in django shell: from django.contrib.auth.models import Group from django.db.models import Count print Group.objects.annotate(cnt=Count('user')).values('id', 'cnt').query.sql_with_params() Django ORM generate