我得到这个错误，当我试图通过蛞蝓在网址可以任何人请协助或如果有任何其他方式来实现这一点。
以下是www.example.com的屏幕截图urls.py|views.py|和models.py
使用Django Rest Framework
Urs.py Screenshot
Views.py Screenshot
Models.py Screenshot
Error Which I am Getting!!
救命啊！
先谢了。
如果屏幕截图不起作用，下面是视图代码

def blog(request):
    blog = Blog.objects.all()
    serializer = BlogSerializer(blog, many=True )
    return Response(serializer.data)

@api_view('GET')
def blogPost(request, slug):
    blogpost = Blog.objects.get(slug=slug)
    serializer = BlogSerializer(blogpost, many=False)
    return Response(serializer.data)

下面是www.example.com的代码urls.py

from django.urls import path
from . import views

urlpatterns = [
    path('blog', views.blog, name='blog'),
    path('blog/<str:slug>/', views.blogPost, name='blogpost')
]

下面是www.example.com的代码models.py

class Blog(models.Model):
    sno = models.AutoField(primary_key=True)
    title = models.CharField(max_length=255)
    slug = models.SlugField(unique=True,  allow_unicode=True, blank=True)
    author = models.CharField(max_length=255)
    read_time = models.IntegerField()
    publish_time = models.DateTimeField(default=timezone.now)
    hero_image = models.ImageField(blank=True,)
    content = models.TextField()
    tags = models.CharField(max_length=100)
    status = models.BooleanField(default=True)
    category = models.CharField(max_length=100)

    def save(self,*args,**kwargs):
        self.slug=slugify(self.title)
        super(Blog,self).save(*args,**kwargs)
    
    def __str__(self):
        return self.title