In Django, how does one filter a QuerySet with dynamic field lookups?


Given a class:

from django.db import models

class Person(models.Model):
    name = models.CharField(max_length=20)

Is it possible, and if so how, to have a QuerySet that filters based on dynamic arguments? For example:

 # Instead of:
 # ... and:

 # ... is there some way, given:
 filter_by = '{0}__{1}'.format('name', 'startswith')
 filter_value = 'B'

 # ... that you can run the equivalent of this?
 # ... which will throw an exception, since `filter_by` is not
 # an attribute of `Person`.
Asked By: Brian M. Hunt



A really complex search forms usually indicates that a simpler model is trying to dig it’s way out.

How, exactly, do you expect to get the values for the column name and operation?
Where do you get the values of 'name' an 'startswith'?

 filter_by = '%s__%s' % ('name', 'startswith')
  1. A “search” form? You’re going to — what? — pick the name from a list of names? Pick the operation from a list of operations? While open-ended, most people find this confusing and hard-to-use.

    How many columns have such filters? 6? 12? 18?

    • A few? A complex pick-list doesn’t make sense. A few fields and a few if-statements make sense.
    • A large number? Your model doesn’t sound right. It sounds like the “field” is actually a key to a row in another table, not a column.
  2. Specific filter buttons. Wait… That’s the way the Django admin works. Specific filters are turned into buttons. And the same analysis as above applies. A few filters make sense. A large number of filters usually means a kind of first normal form violation.

A lot of similar fields often means there should have been more rows and fewer fields.

Answered By: S.Lott

Python’s argument expansion may be used to solve this problem:

kwargs = {
    '{0}__{1}'.format('name', 'startswith'): 'A',
    '{0}__{1}'.format('name', 'endswith'): 'Z'


This is a very common and useful Python idiom.

Answered By: Daniel Naab

A simplified example:

In a Django survey app, I wanted an HTML select list showing registered users. But because we have 5000 registered users, I needed a way to filter that list based on query criteria (such as just people who completed a certain workshop). In order for the survey element to be re-usable, I needed for the person creating the survey question to be able to attach those criteria to that question (don’t want to hard-code the query into the app).

The solution I came up with isn’t 100% user friendly (requires help from a tech person to create the query) but it does solve the problem. When creating the question, the editor can enter a dictionary into a custom field, e.g.:


That string is stored in the database. In the view code, it comes back in as self.question.custom_query . The value of that is a string that looks like a dictionary. We turn it back into a real dictionary with eval() and then stuff it into the queryset with **kwargs:

kwargs = eval(self.question.custom_query)
user_list = User.objects.filter(**kwargs).order_by("last_name")   
Answered By: shacker

Django.db.models.Q is exactly what you want in a Django way.

Answered By: Brent81

Additionally to extend on previous answer that made some requests for further code elements I am adding some working code that I am using
in my code with Q. Let’s say that I in my request it is possible to have or not filter on fields like:


Those fields can appear in query but they may also be missed.

This is how I am building filters based on those fields on an aggregated query that cannot be further filtered after the initial queryset execution:

# prepare filters to apply to queryset
filters = {}
if publisher_id:
    filters['publisher_id'] = publisher_id
if date_from:
    filters['metric_date__gte'] = date_from
if date_until:
    filters['metric_date__lte'] = date_until

filter_q = Q(**filters)

queryset = Something.objects.filter(filter_q)...

Hope this helps since I’ve spent quite some time to dig this up.


As an additional benefit, you can use lists too. For previous example, if instead of publisher_id you have a list called publisher_ids, than you could use this piece of code:

if publisher_ids:
    filters['publisher_id__in'] = publisher_ids
Answered By: Branko Radojevic

This looks much more understandable to me:

kwargs = {
    'name__startswith': 'A',
    'name__endswith': 'Z',
    ***(Add more filters here)***


Answered By: amarse
Categories: questions Tags: , ,
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