[Fixed]-Django queryset: Exclude list of emails using endswith

by

35👍

✅

Simply loop over the QuerySet, as QuerySets are lazy.

emails_to_exclude = ['@example.com', '@test.com', '@mailinator.com' ....]
users = Users.objects
for exclude_email in emails_to_exclude:
    users = users.exclude(email__endswith=exclude_email)
users = users.all()
👤Ahsan

5👍

You can also do this with regular expressions in single query.

emails_to_exclude = ['@example.com', '@test.com', '@mailinator.com' ....]
User.objects.exclude(email__regex = "|".join(emails_to_exclude))

I don’t know the efficiency of this query.

This will not work for SQLite, as it has no built in regular expression support.

👤Zubair Afzal

2👍

One more way to achieve this:

from django.contrib.auth.models import User
from django.db.models import Q

emails_to_exclude = ['@example.com', '@test.com', '@mailinator.com']

users = User.objects.all()

filters = Q()
for ending in emails_to_exclude:
    filters |= Q(email__endswith=ending)

filtered_users = users.filter(~filters)
👤ettanany

1👍

You can probably loop over the emails and build up a Q Object. Actually, you can probably do a 1-liner if you’re clever.

User.objects.exclude(bitwise_or_function[Q(email__endswith=e) for e in emails_to_exclude])

Something like that. I don’t remember the function to bitwise-OR a whole list together, my Python’s rusty.

👤mpen

1👍

This should works with the latest version of Python and Django. The reduce function is a good friend.

from functools import reduce
from operator import or_
from django.db.models import Q

emails_to_exclude = ['@example.com', '@test.com', '@mailinator.com' ....]
users = ( Users.objects
    .exclude( reduce( or_, ( 
        Q(( "email__endswith", k ))
        for k in emails_to_exclude 
    ) ) )
)
👤Sphynx-HenryAY

-1👍

I changed the exclusion input to make it a set and to not have the “@”. Otherwise, this should do what you want.

>>> emails = ['foo@example.com', 'spam@stackoverflow.com', 'bad@test.com']
>>> excludes = {'example.com', 'test.com', 'mailinator.com'}
>>> [email for email in emails if email.split('@')[-1] not in excludes]
['spam@stackoverflow.com']
👤robert

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