Question

我正在尝试创建基于类的详细视图，可以通过单击列表视图项来访问它。问题是在基于函数的视图中很容易实现，但是在基于类的视图中却不能做到这一点。

model.py

func main() {
    http.HandleFunc("/", emailListHandler)
    fs := http.FileServer(http.Dir("./static"))
    http.Handle("/", fs)
    fmt.Println(http.ListenAndServe(":8080", nil))
}

views.py

from django.db import models
import datetime
  # Create your models here.
   class BlogPost(models.Model):


title = models.CharField(max_length=500)
writer = models.CharField(max_length=150,default='my dept')
category =models.CharField(max_length=150)
image = models.ImageField(upload_to='images')
post   = models.TextField(max_length=2000)
Date  = models.DateField( default=datetime.date.today)


def __str__(self):
    return self.title

urls.py

     from.models import BlogPost , EDITORIAL_RESPONSIBILITIES , Reviewers ,Confrences 
             ,ABSTRACT_IN_CONFERENCES
    class BlogList(ListView):
       model = BlogPost
      template_name = 'blog/bloglist.html'
         context_object_name = 'post'
   class BlogDetail(DetailView):
       model = BlogPost
          template_name = 'blogdetail.html'

listview模板运行完全正常。.目录结构也很好。.listviw.html和detail.html都位于template / blog / ..的同一文件夹中。

listview模板

path('list', BlogList.as_view(), name='list'),
path('(?P<id>\d+)/', BlogDetail.as_view())

Answer 1

确保目录正确

templates
   --blog
     ---bloglist.html
     ---blogdetail.html

添加此

class BlogDetail(DetailView):
   model = BlogPost
   template_name = 'blog/blogdetail.html'

def get_context_data(self, *args, **kwargs):
    context = super(BlogDetail, self).get_context_data(*args, **kwargs)
    post = self.get_object()
    context["post"] = post
    print(post)  #//To check if it's returning the right post
    return context

告诉我是否有效

连接列表视图和详细信息视图-django

1 个答案: