Django Restframework POST Object with Array of Objects to One-to-Many tables

Question

I'm new to django and I'm trying to find a solution on how to post an object with array of object inside my database.

So far, here's my sample model.

class Fruit(models.Model):
    name = models.CharField(max_length=100)


class FruitReviews(models.Model):
    reviews = models.CharField(max_length=100)
    fruit = models.ForeignKey(Fruit, on_delete=models.CASCADE)

Here's my serializer.

class FruitSerializer(serializers.ModelSerializer):
    class Meta:
        model = Fruit
        fields = '__all__'

class FruitReviewSerializer(serializers.ModelSerializer):
    class Meta:
        model = FruitReviews
        fields = '__all__'

And Here's my viewset that I'm trying but unfortunately, it doesn't work.

@api_view(['POST'])
def createFruit(request):
    serializer = FruitSerializer(data=request.data)
    serializer2 = FruitReviewSerializer(data=request.data.fruitReviews, many = True)
    if serializer.is_valid() and serializer2.is_valid():
        serializer.save()
        serializer2.save()
        return Response(status=status.HTTP_200_OK)

For the URL, I'm using something like this:

from apiapp import views

urlpatterns = [
    path('fruit-api/', views.createFruit),
]

Basically, for instance, from the client side I want to post something like this

{
    "name": "Apple",
    "fruitReviews": [{
        "reviews": "Very yummy"
    }, 
    {
        "reviews": "Healthy"
    },
    {
        "reviews": "Fresh"
    }]
}

I want it to be posted to the fruit table and fruit reviews properly with their relationships as one-to-many intact. Any help please?

That means you want to save apple in fruit table and reviews in fruitReview table in this POST? — ilyasbbu
– ilyasbbu, Commented Oct 6, 2022 at 3:57
That's right @ilyasbbu is it possible in django? I'm completely new to it — Joker Bench
– Joker Bench, Commented Oct 6, 2022 at 4:02
I'm not confident to my def createFruit because I'm not sure if it's possible. — Joker Bench
– Joker Bench, Commented Oct 6, 2022 at 4:03
It is possible in a way i know but that doesnt check for serializer validations! — ilyasbbu
– ilyasbbu, Commented Oct 6, 2022 at 4:03
And as you run this code do u get any errror messages or simply didn't work? — ilyasbbu
– ilyasbbu, Commented Oct 6, 2022 at 4:05

Hashem · Accepted Answer · 2022-10-06 07:23:01Z

1

By default nested serializers are read-only. If you want to support write-operations to a nested serializer field you'll need to create create() and/or update() methods in order to explicitly specify how the child relationships should be saved:

class TrackSerializer(serializers.ModelSerializer):
    class Meta:
        model = Track
        fields = ['order', 'title', 'duration']

class AlbumSerializer(serializers.ModelSerializer):
    tracks = TrackSerializer(many=True)

    class Meta:
        model = Album
        fields = ['album_name', 'artist', 'tracks']

    def create(self, validated_data):
        tracks_data = validated_data.pop('tracks')
        album = Album.objects.create(**validated_data)
        for track_data in tracks_data:
            Track.objects.create(album=album, **track_data)
        return album

and data should be like:

data = {
    'album_name': 'The Grey Album',
    'artist': 'Danger Mouse',
    'tracks': [
        {'order': 1, 'title': 'Public Service Announcement', 'duration': 245},
        {'order': 2, 'title': 'What More Can I Say', 'duration': 264},
        {'order': 3, 'title': 'Encore', 'duration': 159},
    ],
}

reference: https://www.django-rest-framework.org/api-guide/relations/

answered Oct 6, 2022 at 7:23

Hashem

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2 Comments

Joker Bench Over a year ago

Hello, what's the usage of that? I'm trying to use that but when I'm calling the if serializer.is_valid() serializer.save(), I'm getting the error that my child validated.pop('history')` is error.

Hashem Over a year ago

Hello, this the best practice to deal with one-to-many nested serializer, you can use it in this way better than what you do, I put Django Rest Framework reference to check how they suggest

Imam Hossain Roni · Accepted Answer · 2022-10-06 08:08:52Z

0

I hope it will work without changing your serializers.

@api_view(['POST'])
@transaction.atomic
def create_fruit(request):
    fruit_reviews_data = request.data['fruitReviews']
    if 'fruitReviews' in request.data.keys():
        request.data.pop('fruitReviews')

    fruit_serializer = FruitSerializer(data=request.data)

    fruit_obj = None
    if fruit_serializer.is_valid(raise_exception=True):
        fruit_obj = Fruit.objects.create(**request.data)

    for fruit_review_data in fruit_reviews_data:
        fruit_review_data['fruit_id'] = fruit_obj.id
        fruit_review_serializer =FruitReviewSerializer(data=fruit_review_data)
        if fruit_review_serializer.is_valid(raise_exception=True):
            fruit_review_serializer.save()
         
    return Response(status=status.HTTP_200_OK)

answered Oct 6, 2022 at 8:08

Imam Hossain Roni

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1 Comment

ilyasbbu Over a year ago

shows error django.db.utils.IntegrityError: NOT NULL constraint failed: mainapp_fruitreviews.fruit_id

Sunderam Dubey · Accepted Answer · 2022-10-06 11:42:07Z

0

First you need to change your serializer for FruitReviewSerializer as:

  class FruitReviewSerializer(serializers.ModelSerializer):
      class Meta:
          model = FruitReviews
          fields = ['reviews']

Then in views.py:

  @api_view(['POST'])
  def createFruit(request):
      serializer = FruitSerializer(data=request.data)
      serializer2 = FruitReviewSerializer(data=request.data['fruitReviews'], many = True)
      if serializer.is_valid() and serializer2.is_valid():
          fruit = Fruit.objects.filter(name = request.data['name']).exists()
          if not fruit:
              serializer.save()
          fruit = Fruit.objects.get(name = request.data['name'])
          for i in request.data['fruitReviews']:
              review = FruitReviews(
                  fruit = fruit,
                  reviews = i['reviews']
              )
              review.save()
          return Response(status=status.HTTP_200_OK)
      else:
          return Response(status=400)

edited Oct 6, 2022 at 11:42

Sunderam Dubey

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answered Oct 6, 2022 at 5:50

ilyasbbu

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1 Comment

Joker Bench Over a year ago

Thank you, I upvote this answer because it's also helpful but I accepted the answer Hashem because it's connected to the restframework. Thank you!

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