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 vote354 views1 answer
Django multi column index
Python Django
1 vote339 views1 answer
How to redirect to previous page in django
Python Django
0 vote308 views1 answer
Cant use foreign key in list_display django
Python Django
1 vote356 views1 answer
How to create staff account in django
Python Django
0 vote341 views1 answer
Django pointfield value format for fixtures
Python Django
16 votes355 views6 answers
Django models
Python Django
0 vote322 views1 answer
Manytomany django add bulk create
Python Django
0 vote316 views1 answer
How to use django-rest-framework-datatables
Python Django
3 votes327 views2 answers
Change django administration text
Python Django
1 vote338 views1 answer
What is values_list in django orm
Python Django