I have a raw SQL:

SELECT table1.field1,

(SELECT GROUP_CONCAT(commaSeparatedField SEPARATOR ', ')
FROM table2 WHERE table2.field12=field12) as commaSeparatedField

FROM table1

How do I do this in Django ORM ?

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Two thoughts come to mind here:

  1. Use a raw sql query.

  2. Take a step back and look at why you’re doing this - what the fundamental objective is for the output from this query, then re-evaluate your approach based on Django’s model objects and the ORM to determine what the “Django-style approach” would be within the context of your application’s models.

Not every solution is a direct translation from SQL to the ORM. It’s kinda like translating languages - not every language idiom is a word-for-work conversion. You want to focus on the meaning of what you’re trying to say.

(The “five whys” approach may be useful in cases like this.)

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The following should do

from django.db.models import Aggregate, CharField

# From
class GroupConcat(Aggregate):
    function = 'GROUP_CONCAT'
    template = '%(function)s(%(distinct)s%(expressions)s%(ordering)s%(separator)s)'

    def __init__(self, expression, distinct=False, ordering=None, separator=',', **extra):
            distinct='DISTINCT ' if distinct else '',
            ordering=' ORDER BY %s' % ordering if ordering is not None else '',
            separator=' SEPARATOR "%s"' % separator,


Thanks for the response - I think I should mention the actual use case. There is a table with each row containing many private and public IP addresses - and I want to display the rows with each row showing Public and Private IP addresses concatenated with commas as a single string for private IP address and as a single string for public IP address. These IP addresses are stored in 2 separate tables.

thank you @charettes seems to work well!

can be improved slightly by adding:

class GroupConcat(Aggregate):
    function = 'GROUP_CONCAT'
    template = '%(function)s(%(distinct)s%(expressions)s%(ordering)s%(separator)s)'
    allow_distinct = True
1 Like