Python Django REST 调用返回对象编号而不是对象名称

Question

我是 Python 新手，我想我的序列化不正确。

这是 REST 调用结果:

authors/models.py:

    from django.db import models
from django.contrib.auth.models import User


# Create your models here.

class Author(models.Model):
    name = models.CharField(max_length=100)
    def __str__(self):
        return self.name


class Book(models.Model):
    auto_increment_id = models.AutoField(primary_key=True)
    name = models.CharField('Book name', max_length=100)
    author = models.ForeignKey(Author, blank=False, null=False, related_name='book_author')
    contents = models.TextField('Contents', blank=False, null=False)
    def __str__(self):
        return self.name

authors/views.py

from authors.models import Book, Author
from authors.serializers import BookSerializer, AuthorSerializer
from django.shortcuts import render
from rest_framework import generics


class ListCreateBooks(generics.ListCreateAPIView):
    queryset = Book.objects.all()
    serializer_class = BookSerializer

class ListCreateAuthor(generics.ListCreateAPIView):
    queryset = Author.objects.all()
    serializer_class = AuthorSerializer

authors/serializers.py

from authors.models import Book, Author
from rest_framework import serializers

class BookSerializer(serializers.ModelSerializer):
    class Meta:
        model = Book
        fields = ('name', 'author')

class AuthorSerializer(serializers.ModelSerializer):
    class Meta:
        model = Author
        book_author = 'name'

我是 Django 新手，但我尝试了很多东西，在我的views.py 中，我添加了另一个名为 AuthorSerializer 的类，该类从我在序列化器中创建的相应类导入，但后来我意识到我不知道如何将我的 ListCreateAuthor 添加到:

 url(r'^api-auth/', ListCreateBooks.as_view(), name='list_books')

我用 ListCreateAuthor.as_view() 添加了另一个参数，这给了我一个立即错误(而且也没有多大意义)我在这里走错路了吗？我该怎么办？解决这个问题吗？

EDIT: @Abdulafaja gave a partial answer, which did solve the read, but now after checking a POST insert - it gives an error for create or update.

在 Django rest_framework's documentation (由@Abdulafaja提供的链接)它只给出了嵌套序列化器的一个示例，并且它是一对多关系(专辑->轨道)，但我的是一对一关系(书籍->作者)，所以我不知道如何序列化嵌套特征。该API需要为前端提供所有CRUD功能。

谢谢。

EDITED serializers.py by Abdullah:

from authors.models import Book, Author
from rest_framework import serializers


class AuthorSerializer(serializers.ModelSerializer):
    class Meta:
        model = Author
        book_author = 'name'


class BookSerializer(serializers.ModelSerializer):
    author = AuthorSerializer(many=False)

    class Meta:
        model = Book
        fields = ('name', 'author')

        def create(self, validated_data):
            author, _ = Author.objects.get_or_create(name=validated_data.get('author').get('name'))
            return Book.objects.create(name=validated_data.get('name'), author=author)

        def update(self, instance, validated_data):
            author = validated_data.get('author')
            if author:
                instance.author = Author.objects.get_or_create(name=author.get('name'))
            instance.name = validated_data.get('name', instance.name)
            instance.save()
            return instance

Answer 1

据我了解，您希望在 api 请求中获取作者的详细信息，而不仅仅是他们的编号？

做到这一点的方法是在模型序列化器中设置深度。

class BookSerializer(serializers.ModelSerializer):
    class Meta:
        model = Book
        fields = ('name', 'author')
        depth=1

深度选项应设置为一个整数值，该值指示在恢复为平面表示之前应遍历的关系的深度。即 1、2、3...

来源:http://www.django-rest-framework.org/api-guide/serializers/