how to store a list of images in postgreSQL ArrayField (django rest)

Question

I'm developing a project using django-rest for my backend and postgreSQL as my database.

I'm using postgreSQL ArrayField in my Recipe model to store multiple images of different instructions.

However when I send a POST request with a list of instructions images (instructions_image_list) I'am getting a weird error.

I would really appreciate if anyone could help me solve it. thanks in advance.

the error

status_code 500
'utf-8' codec can't decode byte 0xff in position 0: invalid start byte

the model

class Recipe(models.Model):
    id = models.UUIDField( 
        primary_key=True,
        default=uuid.uuid4,
        editable=False,
    )
    author = models.ForeignKey(get_user_model() , on_delete=models.CASCADE, null=True, related_name='author') 
    photo_main = models.ImageField(upload_to='media/', blank=True)
    title = models.CharField(max_length=150)
    instructions_text_list = ArrayField(models.CharField(max_length=100, blank=True,null=True),default=list, blank=True ,null=True,size=15,)
    instructions_image_list = ArrayField(models.ImageField(upload_to='media/',blank=True, null=True),default=list,  blank=True ,null=True, size=15,)

the view

class RecipeCreate(CreateAPIView):
    permission_classes = (permissions.IsAuthenticated, )
    queryset = Recipe.objects.all()
    serializer_class = RecipeCreateSerializer

    def perform_create(self, serializer):
        '''save the the current logged in user as the author of the recipe'''
        serializer.save(author=self.request.user)

Arjun Shahi · Accepted Answer · 2021-12-29 11:21:26Z

1

It's better if you store recipe images in separate table instead of using ArrayField.

class Recipe(models.Model):
......

class RecipeImage(models.Model):
   recipe = models.ForeignKey(Recipe, on_delete=models.CASCADE)
   image = models.ImageField(upload_to="recipe_images/")

EDIT:

You can handle your multiple recipe images within your RecipeCreateSerializer and RecipeCreateView.

For example:


# serializer

class RecipeCreateSerializer(serializers.ModelSerializer):
      recipe_images = serializers.ListField(
        child=serializers.FileField(allow_empty_file=False, use_url=False)
    )

      class Meta:
        model = Recipe
        fields = [
           #  .. other fields
            "recipe_images",

        ]

      def create(self, validated_data):
         validated_data.pop("recipe_images", None)
         return super().create(validated_data)

# view

from django.db import transaction

class RecipeCreate(CreateAPIView):
    permission_classes = (permissions.IsAuthenticated, )
    queryset = Recipe.objects.all()
    serializer_class = RecipeCreateSerializer
    
    @transaction.atomic
    def perform_create(self, serializer):
        '''save the the current logged in user as the author of the recipe'''
        recipe = serializer.save(author=self.request.user)
        # create recipe images here with each images uploaded
        for img in self.request.data.getlist("recipe_images"):
             RecipeImage.objects.create(image=img, recipe=recipe)
        return super().perform_create(serializer)

edited Dec 29, 2021 at 11:21

answered Dec 28, 2021 at 15:51

Arjun Shahi

7,3404 gold badges14 silver badges31 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

EliyaMelamed Over a year ago

Thanks for the help, I thought about using this method but it's get way more messy than saving the images in ArrayField, is there any way to make my method work?

Arjun Shahi Over a year ago

How's it way more messy it should not be messy as it's the standard way for your requirement.

EliyaMelamed Over a year ago

Cause I will need to create manyToManyField + create different view to create the RecipeImage model + adding permissions + assigning the created models to the Recipe manyToManyField

Arjun Shahi Over a year ago

Why you need many to many relationships? The model i posted in the answer should work and it's not necessary to create another view. You can simply create images within your recipe create view.

EliyaMelamed Over a year ago

1. The manyToManyFiels is because I want to store multiple RecipeImage models in the Recipe model am I wrong? ,2. which field should I send in the post request to assign the images? Thanks in advance

|

Collectives™ on Stack Overflow

how to store a list of images in postgreSQL ArrayField (django rest)

1 Answer 1

8 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

8 Comments

Your Answer

Sign up or log in

Post as a guest

Related