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

0 vote298 views1 answer
Django float validator
Python Django
0 vote213 views1 answer
Django user refistration
Python Django
0 vote200 views1 answer
Django swagger settings https
Python Django
0 vote238 views1 answer
How i dynamically populate to CORS_ALLOWED_ORIGINS in django rest framework
Python Django
0 vote199 views1 answer
How to add phone number to django user model
Python Django
-1 vote200 views1 answer
Binascii.Error: Incorrect padding in python django
Python Django
0 vote269 views1 answer
Django admin.py all fields
Python Django
-2 votes193 views4 answers
Django datepicker
Python Django
0 vote200 views1 answer
Django create username and password from csv
Python Django
0 vote202 views1 answer
Django insert bulk data
Python Django