How to get unique values with id in django from database

Question

So this is my query and it displays only file name but I want 'id' also.

filenames=Upload.objects.values('file').distinct().filter(created_by=request.user).all()

this will generate only unique files as dict. But I want 'id' of those files also and if use 'id' inside that 'values' it will not give unique file names. So this is my issue. I'm okay with other methods as well to solve this issue.

Are you using Postgresql for your application? And there are something you have to follow when you want to query distinct data on specific field, https://docs.djangoproject.com/en/dev/ref/models/querysets/#distinct . Make sure that you added `order_by` into your query code — Toan Quoc Ho, Sep 25 '19 at 06:35
Possible duplicate of [Django distinct group by query on two fields](https://stackoverflow.com/questions/33088337/django-distinct-group-by-query-on-two-fields) — wfehr, Sep 25 '19 at 06:35
@wfehr if use query like that, then it will not give unique values. — Sharath Nayak, Sep 25 '19 at 06:39

Yugandhar Chaudhari · Answer 1 · 2019-09-25T06:55:43.093

0

This will work

from django.db.models import F

Upload.objects.filter(created_by=request.user).values('file').distinct().annotate(id=F('pk'))

edited Sep 25 '19 at 06:55

answered Sep 25 '19 at 06:39

Yugandhar Chaudhari

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Sure. Will try it out. – Sharath Nayak Sep 25 '19 at 07:02
I'm getting all values. – Sharath Nayak Sep 25 '19 at 08:39

score -1 · Answer 2 · answered Sep 25 '19 at 09:12

I don't think we can achieve it using Distinct.

So I used loop like this.

filenames=Upload.objects.filter(created_by=request.user).all()
seen = set()
result = []
for f in filenames:
    if f.file not in seen:
        result.append(f)
        seen.add(f.file)
print result

How to get unique values with id in django from database

2 Answers2