Django Rest框架:通过slug而不是ID访问项目详细信息 [英] Django Rest Framework: Access item detail by slug instead of ID
问题描述
是否可以使用对象的插槽(或任何其他字段)访问项目的详细信息,而不是使用ID?
例如,如果我有一个项目与sluglorem和ID 1.默认情况下,URL是 http:// localhost:9999 / items / 1 /
。我想通过 http:// localhost:9999 / items / lorem /
访问它。
添加<序列化器的Meta类中的code> lookup_field 没有改变自动生成的URL,也没有允许我通过手动编写slug而不是URL中的ID来访问该项。 p>
models.py
class Item(models.Model):
slug = models.CharField(max_length = 100,unique = True)
title = models.CharField(max_length = 100,blank = True,default ='')
# URL
item_url = models.URLField(unique = True)
serializers.py
class ClassItemSerializer(serializers.HyperlinkedModelSerializer):
class Meta:
model = Item
fields =( 'url','slug','title','item_url')
views.py
class ItemViewSet(viewsets.ModelV iewSet):
queryset = Item.objects.all()
serializer_class = ItemSerializer
urls.py
router = DefaultRouter()
router.register(r'items',views.ItemViewSet )
urlpatterns = [
url(r'^',include(router.urls)),
]
生成的JSON:
[
{
url:http:// localhost:9999 / items / 1 /,
slug:lorem,
title:Lorem,
item_url:http://example.com
}
]
您应该在序列化程序中设置 lookup_field
:
class ItemSerializer(serializers.HyperlinkedModelSerializer):
class Meta:
model = Item
fields =('url','slug','title','item_url' )
lookup_field ='slug'
extra_kwargs = {
'url':{'lookup_field':'slug'}
}
您的观点:
类ItemViewSet(viewsets.ModelViewSet):
queryset = Item.objects.all()
serializer_class = ItemSerializer
lookup_field ='slug'
我得到这个结果: p>
〜curl http://127.0.0.1:8000/items/testslug/ | python -mjson.tool
{
item_url:https://example.com/,
slug:testslug,
title:测试标题,
url:http://127.0.0.1:8000/items/testslug/
}
Is it possible to use a object's slug (or any other field) to access the details of an item, instead of using the ID?
For example, if I have an item with the slug "lorem" and ID 1. By default the URL is http://localhost:9999/items/1/
. I want to access it via http://localhost:9999/items/lorem/
instead.
Adding lookup_field
in the serializer's Meta class did nothing to change the automatically generated URL nor did it allow me to access the item by manually writing the slug instead of the ID in the URL.
models.py
class Item(models.Model):
slug = models.CharField(max_length=100, unique=True)
title = models.CharField(max_length=100, blank=True, default='')
# An arbitrary, user provided, URL
item_url = models.URLField(unique=True)
serializers.py
class ClassItemSerializer(serializers.HyperlinkedModelSerializer):
class Meta:
model = Item
fields = ('url', 'slug', 'title', 'item_url')
views.py
class ItemViewSet(viewsets.ModelViewSet):
queryset = Item.objects.all()
serializer_class = ItemSerializer
urls.py
router = DefaultRouter()
router.register(r'items', views.ItemViewSet)
urlpatterns = [
url(r'^', include(router.urls)),
]
Generated JSON:
[
{
"url": "http://localhost:9999/items/1/",
"slug": "lorem",
"title": "Lorem",
"item_url": "http://example.com"
}
]
You should set lookup_field
in your serializer:
class ItemSerializer(serializers.HyperlinkedModelSerializer):
class Meta:
model = Item
fields = ('url', 'slug', 'title', 'item_url')
lookup_field = 'slug'
extra_kwargs = {
'url': {'lookup_field': 'slug'}
}
and in your view:
class ItemViewSet(viewsets.ModelViewSet):
queryset = Item.objects.all()
serializer_class = ItemSerializer
lookup_field = 'slug'
I got this result:
~ curl http://127.0.0.1:8000/items/testslug/ | python -mjson.tool
{
"item_url": "https://example.com/",
"slug": "testslug",
"title": "Test Title",
"url": "http://127.0.0.1:8000/items/testslug/"
}
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