Django group by

# If you mean to do aggregation you can use the aggregation features of the ORM:
from django.db.models import Count
Members.objects.values('designation').annotate(dcount=Count('designation'))

# This results in a query similar to:
SELECT designation, COUNT(designation) AS dcount
FROM members GROUP BY designation

#and the output would be of the form
[{'designation': 'Salesman', 'dcount': 2}, 
 {'designation': 'Manager', 'dcount': 2}]

Source: stackoverflow.com

Similar codes

1 vote224 views1 answer
Radiobuttons django
Python Django
3 votes273 views1 answer
Django user group check
Python Django
1 vote267 views2 answers
How to access cookies in django
Python Django
0 vote221 views1 answer
Django admin.py date format
Python Django
0 vote210 views1 answer
Slug in redirect django
Python Django
0 vote234 views1 answer
Request.build_absolute_uri django
Python Django
0 vote243 views1 answer
Create fixtures django
Python Django
0 vote248 views1 answer
Iniciar servidor en Django
Python Django
0 vote270 views1 answer
How i dynamically populate to CORS_ALLOWED_ORIGINS in django rest framework
Python Django
0 vote219 views1 answer
How to add value in array django
Python Django