Django referencing a specific object in a many-to-one relationship

Question

Let's say I have the following models:

from django.db import models

class Reporter(models.Model):
    first_name = models.CharField(max_length=30)
    last_name = models.CharField(max_length=30)
    email = models.EmailField()

class Article(models.Model):
    headline = models.CharField(max_length=100)
    pub_date = models.DateField()
    reporter = models.ForeignKey(Reporter, related_name="articles", on_delete=models.CASCADE)

I'd like to add a favorite_article field to my Reporter model that will reference a specific Article from reporter.articles.

One option is put the information into the Article model instead:

class Article(models.Model):
    headline = models.CharField(max_length=100)
    pub_date = models.DateField()
    reporter = models.ForeignKey(Reporter, related_name="articles", on_delete=models.CASCADE)
    is_favorite = models.BooleanField()

But this doesn't seem like a very clean solution. Is there a better method to do this?

Have you tried anything yet or do you have any ideas on how to go about this? — Henry Woody, Apr 15 '21 at 16:51
@HenryWoody I added a possible method that I've considered. I'm hoping that there is a cleaner/better solution. — USS1994, Apr 15 '21 at 17:02

Henry Woody · Accepted Answer · 2021-05-04T23:45:11.110

The approach you've suggested will work, however in its current form it allows for multiple Articles to be the favorite of one Reporter. With a bit of extra processing you can ensure that only one (at most) Article per Reporter is the favorite.

Making a few modifications to a couple of the answers to the question Unique BooleanField value in Django? we can restrict one True value per Reporter rather than one True value for the entire Article model. The approach is to check for other favorite Articles for the same Reporter and set them to not be favorites when saving an instance (rather than using a validation restriction).

I'd also suggest using a single transaction in the save method so that if saving the instance fails the other instances are not modified.

Here's an example:

from django.db import transaction

class Article(models.Model):
    headline = models.CharField(max_length=100)
    pub_date = models.DateField()
    reporter = models.ForeignKey(Reporter, related_name="articles", on_delete=models.CASCADE)
    is_favorite = models.BooleanField(default=False)

    def save(self, *args, **kwargs):
        with transaction.atomic():
            if self.is_favorite:
                reporter_id = self.reporter.id if self.reporter is not None else self.reporter_id
                other_favorites = Article.objects.filter(is_favorite=True, reporter_id=reporter_id)
                if self.pk is not None: # is None when creating a new instance
                    other_favorites.exclude(pk=self.pk)
                other_favorites.update(is_favorite=False)
            return super().save(*args, **kwargs)

I've also changed the approach to use a filter rather than a get just in case.

Then to get the favorite article for a reporter, you can use:

try:
    favorite_article = reporter.articles.get(is_favorite=True)
except Article.DoesNotExist:
    favorite_article = None

which you could wrap into a method/property of the Reporter class.

Thanks @Henry Woody, this works, but I was hoping for something that kept the `favorite_article` field tied to the reporter, so that it can be directly referenced, rather than having to search through (potentially many) articles to find the one marked as the favorite. — USS1994, May 03 '21 at 19:35
@USS1994 You can retrieve the favorited article with `reporter.articles.get(is_favorite=True)` (or just `filter`) so you don't have to search through. Also you should wrap this in a `try`/`except` in case the reporter doesn't have a favorited article. Does that approach work for you? — Henry Woody, May 03 '21 at 21:04

Django referencing a specific object in a many-to-one relationship

1 Answers1