字段名称“get_thumbnail”对于模型“Product”无效

Question

经过几个小时的解决，访问了数百个网站来搜索这个问题。我获得了很多知识，但是却解决了我的问题。

Exception Type: ImproperlyConfigured

Exception Value: Field name get_thumbnail is not valid for model Product.

models.py

from io import BytesIO

from PIL import Image
from django.core.files import File
from django.db import models


class Category(models.Model):
    name = models.CharField(max_length=255)
    slug = models.SlugField()

    class Meta:
        ordering = ('name',)

    def __str__(self):
        return self.name

    def get_absolute_url(self):
        return f'/{self.slug}/'


class Product(models.Model):
    category = models.ForeignKey(Category, related_name='products', on_delete=models.CASCADE)
    name = models.CharField(max_length=255)
    slug = models.SlugField()
    description = models.TextField(blank=True, null=True)
    price = models.DecimalField(max_digits=6, decimal_places=2)
    image = models.ImageField(upload_to='uploads/', blank=True, null=True)
    thumbnail = models.ImageField(upload_to='uploads/', blank=True, null=True)
    date_added = models.DateTimeField(auto_now_add=True)

    class Meta:
        ordering = ('-date_added',)

    def __str__(self):
        return self.name

    def get_absolute_url(self):
        return f'/{self.category.slug}/{self.slug}/'

    def get_image(self):
        if self.image:
            return 'http://127.0.0.1:8000' + self.image.url
        return ''

    def get_thumbnails(self):
        if self.thumbnail:
            return 'http://127.0.0.1:8000' + self.thumbnail.url
        else:
            if self.image:
                self.thumbnail = self.make_thumbnail(self.image)
                self.save()

                return 'http://127.0.0.1:8000' + self.thumbnail.url
            else:
                return ''

    def make_thumbnail(self, image, size=(300, 200)):
        img = Image.open(image)
        img.convert('RGB')
        img.thumbnail(size)

        thumb_io = BytesIO()
        img.save(thumb_io, 'JPEG', quality=85)

        thumbnail = File(thumb_io, name=image.name)

        return thumbnail

serializers.py

from rest_framework import serializers
from .models import Category, Product


class ProductSerializer(serializers.ModelSerializer):
    class Meta:
        model = Product
        fields = (
            "id",
            "name",
            "get_absolute_url",
            "description",
            "price",
            "get_image",
            "get_thumbnail",
        )

views.py

from rest_framework.views import APIView
from rest_framework.response import Response

from .models import Product
from .serializers import ProductSerializer


class LatestProductsList(APIView):
    def get(self, request, format=None):
        products = Product.objects.all()[0:4]
        serializer = ProductSerializer(products, many=True)
        return Response(serializer.data)

Answer 1

也许只是方法名？我看到你将它命名为 get_thumbnails 而不是 get_thumbnail

字段名称“get_thumbnail”对于模型“Product”无效

Field name `get_thumbnail` is not valid for model `Product`

django-rest-framework